Vedic Mathematical Concepts and Their Application to Unsolved Mathematical Problems: Three Proofs of Fermat's Last Theorem

8/20/2019 Vedic Mathematical Concepts and Their Application to Unsolved Mathematical Problems: Three Proofs of Fermat's …

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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 ndi a n I ns t i tu te o f Ma har i s h i V e di c

Sci e nce and Te chnology

Preface

Th e f o l low i ng thre e p roof s , base d on the ap p l i cat i on o f V e di c Ma the m at 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 L a s t T h e o r e m . T h e

f o l low i ng p as sage o f Jaba l i Up ani sha d p rovi de d the s t ructura l k e y f or de v e lop i ng the

mult i d i me ns i onal sp ace s use d i n the argume nt f or F e rmat ' s Las t The ore m:

Ad dr es s co rr es po nd en ce to : S.K. Kapoor, V is it in g Pr ofe ssor

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

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,

"How 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 earl ie r wo rk, The Rationale of О m—Its Formulation, Significance and O c c u r

re nce and Applications in Ancient Literature, I co mm e n te d on the abov e p as sag e as fo l loWS:

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.


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MODERN SCIENCE AND VEDIC SCIENCE

On this, the Jabali Rishi disclosed that he was enlightened on the points by Shada-

nan. 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 do

main. 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 do

main as 6x3/2 = 9.

Th e signif ica nce of the abo ve with in the ari thme tic doma in is that in orde r to

und er st an d the stru ctur al frames and sy ste ms of natural num ber 9 ( in Upa nis had ic

lan gua ge: B ra hm av idy a) we have to deve lop the under s tand ing in terms of the

st ructura l f rames and system s of natural numbe r 6 ( in Upan isha dic language Sh ada

nan bestowed enl ightenment upon Jabal i Rishi ) . And unders tanding the s t ructural

fra mes and sy ste ms of natur al num be r 6 must be in term s of the structural frame s

and syst em s of natu ral num be r 4 ( in the Upan ish ad ic lan gua ge Ishan be sto we d en

l ightenment upon Shadanan) . This reverse sequent ial process contains the s t ructu

ral ke y.

The basic Vedic Mathematical concepts used are that the unity (single-syl lable

Om) is processable quarter by quarter (Shri Pada processing line) and the fourth

qua rt er is the int egra tio n of the first thr ee quart ers (Maharishi processing line). As

such, I interpret the above Upanishadic passage in the following way: the structural

frames and systems of natural number 4 (and hence the fourth-dimensional domain)

are to be handled as unity, admitting processing quarter by quarter, and the fourth


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VEDIC MATHEMATICAL CONCEPTS

quarter (the unmanifested quarter) is to be processed as integrating the first three

quarters (manifested quarters) . The processing within the fourth quarter along the

North-East diagonal would lead to the domain of Shadanan (admitt ing structuralframes and systems of natural number 6 and hence a six-dimensional domain). The

Upanishadic command is to reverse the process, as the processing is to be had in the

descending order from 9 to 6 and 6 to 4, but here in this world everything is hanging

upside dow n urdh vamu lam— Bhag ava d-Gi ta , 15 .1). This means that

while our aim is to process within the unmanifested quarter in order to go from the

fourth-dimensional domain of natural number 4 to the sixth-dimensional domain of

natural number 6, we have to process along the North-East diagonal but in the direc

t ion which leads from the sixth-dimensional domain towards the fourth-dimensional

domain. Further, the processing must be quarter by quarter , which means that when

natural number 4 is taken as a unity, its quarter would be one, and similarly, when

natural number 6 is taken as a unity, its processing unit would be one. Hence the pro

cessing along the North-East diagonal is to be had by dividing it into six parts. Out of

these six parts, only one part would be unmanifested and the remaining five parts

would cover the manifested part. With this the procedure of structural processing

stands as fol lows:

Step 1: Th e proc es sin g is to be in the fourth quarter of the Om form ulat ion.

Step 2: Fourth quarter (fourth com pon ent ) of the Devana gari Om formulation is like

a two-dimensional cartesian frame.

Step 3: So processing within the fourth quarter (fourth component) amounts to pro

cessing on the format of a plane.

Step 4: A s bot h the dime nsi ona l lines are symm et ri c, the natur al figure of the pla ne

format is a square (to be called the processing square).

Step 5: F or pro ce ss ing of the first thre e qua rter s of the manife ste d dom ai ns , divide

the square into 3x3 = 9 squares (to be called processing units).

Step 6: Therefore, if the length of the square is Z, then the length of the processing

unit would be Z/3.

Step 7: As the fourth quart er is to be pro ce ss ed as the integ ratio n of the first three

quarters, the length of the processing square should be increased by the length of its pro

cessing units, that is, the extended processing square, as it may be called, would have a

length equal to Z + Z/3 = 4Z/3.

Step 8: Th e proce ssing squ are being of length Z, the manifeste d part of the Nort h-

East diag ona l of the extende d proc ess ing square would be 5Z/6 , and the unma nifes ted

part of the North-East diagonal of the said square would be Z/6.

Step 9: The above format develops a square whose length is equal to the diagonal,

which means we are transcending from the geometrical square to the values-square (as

it may be called). In this format transcendence is permissible within the format of a geo

met ric al sq ua re itself in ter ms of the cro ss po ints of the lines paralle l to the ax es of the

two -di men sio nal cartes ian frame of the geomet rica l square . If we take into accou nt only

the cross points of the lines parallel to the axes (which lines may be called parallel axes)

the n the cros s poi nts on the length or bre ad th of the squa re woul d be equa l to the cro ss

poi nts on any of the two dia gona ls of the squ are.


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Step 10: T he effect of ext end ing the pro ce ssi ng squa re into the extend ed pro ces sin g

square upon the values-square of the processing square qua the values-square of the ex

ten ded proc es sin g squ are is the structura l key for reso lving why the sum of tw o regular

bodies of Nth-dimensional space does not constitute another regular body of the Nth-

dimensional space.

In the present studies I have applied the above structural key in three different ways

to prove Fermat's Last Theorem. These proofs are submitted not only for the purpose

of supplying a proof of this unsolved theorem, but also with the aim of appealing to

other scholars to approach the main challenges within their disciplines through the Vedic

wisdom.

The Vedic perspective on methodology integrates objectivity with subjective experi

ence and as such those who are trained in objective methodology are required only to

learn h ow to sup ple me nt this appr oac h wit h their ow n subjective experi ence of the Ved ic

Reality. For this I feel highly privileged to be at the feet of His Holiness Shri Pada Ba-

baji who initiated me for the Shri Pada processing line to process this quarter by quarter,

and at the feet of Hi s Hol ine ss Maha ris hi Ma he sh Yogi who initiated me for the Mah ar i-

shi pro ce ss ing line to pro ce ss the fourth quar ter as the integra tion of the first three quar

ters. I am also highly obliged to Professor Krishnaji, Chairman, Indian Institute of Ma-

harishi Ved ic Scie nce and Tec hnol ogy, for experi enced guida nce and persona l interest

in these 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 number.

Step 5: Hence, in the equation UN = V N + T N , V and T are irrational numbers. There

fore , we have to look for solu tions of this equa tion in the field of real nu mb er s. Her e it

may be relevant to note that U N , V N , and T N are rational numbers, so U 1 = UN , V 1 , = V

N

and T1 = TN give rise to the equation U 1 = V1 + T 1 where U 1, V1 and T1 are rational num

be rs . Th is has the obv iou s solut ion in the field of rationa l num be rs . Howe ve r, whe n the

solution is required for UN

= VN

+ TN

, where V and T are irrational numbers, we have to

shift to the field of rea l num be rs .

Step 6: Th e field of real num ber s constitut es the arithmetical cont inuu m w hich is

equivalent to the linear geometrical continuum of a straight line. Now the solutions of

the equations UN

= VN

+ TN

and MN .

UN

= MN .

VN

+ MN .

TN

are directly linked up. There

fore, the solution of the equation ( MU )N

= (MV)N

+ (MT)N

, which is nothing but SN

=


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point representing the number zero (0). As a second step, for a given rational num

be r Z, we ma y cut a cl ose d inte rva l [O Z] of len gth OZ = Z from the line OR. As a

third step, for any rational number X < Z, we can cut a closed interval [OX] ofle ng th OX = X from the inte rva l [OZ ] of len gth OZ = Z.

4. The interval [OZ] gets divided into two parts, namely closed interval [OX] and

the one-sided open interval (XZ]. Both the parts, namely [OX] and (XZ], do not

have any common point. All numbers rational and irrational less than or equal to X

are in [OX], and all numbers rational or irrational greater than X are in (XZ]. As

such, the interval (XZ] is of an irrational length.

Fermat's Last Theorem

5. Statement: "It is impossible to separate a cube into two cubes, or a biquadrate

into two biquadrates, or in general any power higher than the second into powers of

l ike degree."— Fermat

6. In modern mathematical language this is restated as: xn

+ yn

= zn

admits a solution

in natural numbers only for n = 2. That means for given natural numbers z, n, and x,

with x < z, we cannot find a natural number whose nth degree is equal to z n - x n .

Proof

7. Let V = z n - x n .

8. z, n, and x are natural numbers, therefore z n and x n are also natural numbers.

Le t Z = zn

and X = xn

. Therefore, V = Z-X.

9. As Z and X are natu ral nu mb er s, so the clos ed inte rva l [OZ] of le ngt h Z cu ts a

rat iona l length on the real line OR. Si mil arl y, the clos ed inter val [O X] of leng th X

cuts a rational length from the interval [OZ].

10. V = Z - X = [O Z ]- [O X] = (X Z] = irrat ional length.

11 . H e n c e , V N" for every natural num ber N, as N n is a natural number so

would equal a rational length and not an irrational length.

12. Th e ab ov e pr ov es the th eo re m but for the res tri cti on for the n to be 3.

Restrictions for n to be 3

Rationale for the restr ict ions:13 . To arr ive at the rati ona le for the rest ric tio ns for n to be 3, we ha ve to go

ba ck to our defini tion of the dime ns ion in ter ms of its geo me tri ca l rep res ent ati on.

14. Th e defini t ion of dim ens ion can not be expr esse d except within the manifes t

ed wor ld of thre e-d ime nsio nal objects, ther eby permit t ing its arr ang eme nt as a l ine

ar geo me tri ca l con tin uum of straight l ines, each a math ema tica l model of the ari th

met ic al cont in uum of real numb ers . In concrete terms, geometr y accepts d imensi on

as that which completely permits representat ion as a straight l ine, and the geometri

cal universe fol lowing is the three-dimensional space with the three dimensions

nothing but straight lines. In fact, here lies the rationale and the answer to the ques

t ion why restr i ct ions are plac ed upon n to be 3. How eve r, com pre he nsi on of the

ma nif es ted worl d can not be ac cep ted as a proof as such of the exis ten ce or othe r

wise of the higher- dimensional spaces. For this we have to have a purely mathemat

ical approach.

15. For this, we may have insight into the internal structure of N n by accepting it


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MO DE RN SCIENCE AND VEDIC SCIENCE

to be a regul ar bo dy of len gth N of nt h di me ns ion al sp ac e (for the presen t take as a

de f i ni t i on that de gre e n re p r e se n ts the nu mb e r o f d i m e ns i on s) .

The First Dimensional Regular Body

16. To have p rop e r i ns i ght i nto the i nte rna l s t ructure o f h i gh e rd i me ns i o nal regu-

lar bodies, we would be required to pinpoint the rat ionale as to why the very f i rs t

d i m e n s i o n a l r e g u l a r b o d y ( ll) doe s not p e rmi t d i v i s i on i nto smal le r re gu lar bodi e s o f

the f i r s td i me ns i onal sp ace .

17. Th e ans we r i s be ca us e of the very def ini t io n of the natu ral nu m be r 1. Th e nat-

ura l nu mb e r 1 i s the smal l e s t natura l nu mb e r , so the que s t i o n o f the e x i s te n ce o f s t i ll

s m a l l e r n a t u r a l n u m b e r s d o e s n o t a r i s e . H e n c e l1 X

1 + Y

1, where X and Y are natu-

r a l n u m b e r s .

18 . Th e above s i tuat i on de se rv e s p ro p e r ma the ma t i z at i on . F i r s t ly , i t tak e s on e as

unde f i ne d . S i mul tane ous ly i t g i ve s us the f re e dom to acce p t any l i ne ar le ngth, may 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 it an d h e n c e " o n e . "

19. Usi ng the abov e f re edo m of ch oi ce to ac cep t any l inear length as the dimen-

sio nal un it help s us to red uc e any regul ar bo dy N" to the f i rs t regu lar body of nth

di me ns io na l spac e ( Г ) , wh er e N = on e uni t . This also can be take n as the fi rs t regu-

lar bo dy o f the f i r s td i m e ns i on al sp ace ( l1

) s ince 1n

= 1 = l1

.

20. There fore, in order to und erst and the internal s tructural arran gem ent s of dimen-

sional regular bodies, we have no option but to consider the s tructural knot responsible

for transforming unit linear length into a rational fraction plus 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 stru ctural knot of dimen sional re gular

bodies. W e f o c u s o n the s ta te me nt that one l i ne ar uni t is e q u a l to a ha l f uni t ra t i onal

le ngth p lus a ha l f uni t i r ra t i onal le ngth .

22. Before we take up the above s tatement for e laboration, I would l ike to add

her e tha t the rat ion ale for the ab ov e l ies in the f i rst pri nc ipl e (of ma nif est ed worl d

ad mi t t i ng s t ruct ure s ) that the s t ructur a l uni t i s ha l f a s co mp ar e d to one as a d i me n-

s i onal uni t .

Defin ition of a Strai ght Lin e

23 . On e attr ib ute of the s tra ight l ine i s that i t ha s a mi dd le point . A sec on d attr ib-

ute of the s tr aigh t l ine i s that i t ha s a mi n im u m of thr ee po int s . I t may no t be poss i-

b l e to g i v e a p r e c i s e de f in i t i o n of a s t ra i g h t l i ne in t e r m s of s o m e a t t r i b u t e s . How-

ever , the basi c attrib utes of the straigh t line wo ul d help us settle so me pract ical defini-

t i on s . Th e p rac t i ca l de f i ni t i on s ma y have w ort h f or any dom ai n o f p ract i ca l i mpor-

t a n c e b u t t h e s a m e m a y n o t b e a c c e p t a b l e t o m a t h e m a t i c s . A p r e c i s e m a t h e m a t i c a l

de f i n i t i on o f a s t ra i ght l i ne can be p ro vi d e d by the ar i th me t i ca l co nt i n uu m of the re a l

nu mb e r s . Th e f ormat be n e at h the ord e re d d i sp lay o f the set o f re a l nu mb e rs w oul d

mathe mat i ca l ly qua l i f y to be de s i gnate d as a s t ra i ght l i ne . As such th i s i s acce p te d as

the de f i ni t i on o f the s t ra i gh t l i ne f or the p urp ose o f e xp r e s s i o ns o f d i m e ns i on al

f ra me s, as wel l as for the pur po se of cu tt ing len gth s f rom a regu lar bo dy .

Structural Unit Versus Dimensional Unit

2 4 . N o w le t us tak e up the chal le n ge o f the s t ructura l k not (p ar agra p h 2 1) . H e re

the s i tua tio n pr oc ee ds in two step s: f i rs t ly , in te rm s of the an sw er to the poser, i f we


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can transcend the three dimensional geometrical universe and have a geometrical

continuum, and secondly, in terms of the fine connection between the structural unit

and the dimensional unit maintaining the ratio of 1/2 : 1.

Square and Cube

25. For both the steps, initially we may proceed with the structural commonness

of the square as a regular body of a two-dimensional space and the cube as a regular

body of a three-dimensional space.

26. The significance of the structural commonness of a square and a cube is that

the ratio of area and perimeter of the square, and volume and surface area of the

cube, admit sequential order as a function of the degree of the dimensional space.

We may define the degree of n-dimensional space as n. These ratios for length 'a ' of

the regular bodies is:

a 2 : 4a and a 3 : 6a 2

27. The above ratios have the formulation an/2nan-1 : 1 for n = 2 and 3.

28. The above sequential formulation immediately supplies the key to unlock the

structural knot of the dimensional regular body. It is that for any dimensional regular

body the domain part an

(which for the square is the area of the square and for the cube is

its volume) and the frame part 2na n- 1 (which for the square is its perimeter and for the

cube is its surface area) maintain a ratio dependent upon 1/2, the length (a) and the di

mensional degree (n). This ratio comes to be a/2n. The structural significance is that for

any a and any n, there comes 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 de

grees. This factor being free of the length and degree of dimensional regular bodies ac

quires universal application. Hence, even the regular bodies of unit length of all the

dimensional spaces of any degree also admit the factor (1/2). By this means we gain in

sight into the internal structural arrangements of l1

, l2

, 13

. . . 1n

.

Framed Domains Sequence

29 . Befo re we inv esti gate the in tern al struct ure of regu lar bod ies of unit lengt h of

different dimensional spaces, let us understand the sequence [an/2nan-1 fo r n =

1,2,3...]. This sequence may be defined as a framed domains sequence and its indi

vidual terms as individual framed domains or simply framed domains. The reason

for the choice of terminology is that the regular bodies have their domain part con

tained within the frame part and thus can be designated as framed domains.

First Framed Domain

30. The first member of the framed domains sequence [an/2nan-l for n = 1,2,3...],

that is an/2nan- 1 for n = 1, that is a1/2xla1-1, that is a/2, is the first framed domain.

This indicates that out of any closed linear interval we can remove one closed inter

val of half length while the second half would not be a closed interval. Hence the

str uct ura l unit is half of the dime nsi on al unit, whic h essential ly is the linear unit.

One may revert back to this framed domain after having been through the internal


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s t ru c t u re of seco nd and t h i rd f ramed dom ain s, w h ic h a re t h e w el l kno w n geomet ri-

cal figu res of a sq ua re and a cu be , to fully realiz e this im pli cat io n.

Internal Structu re of a Squ are

3 1 . Th e sq u are i s a t w o di men sio na l regu l ar b ody w i t h t h e same l engt h on b ot h

dimensiona l l ines . Taking l engt h t o b e a units, the square would have an area a2 and

a p e r i m e t e r 4a . B y suita bly ch oo si ng a l in ear unit , we can have the len gth of a

square equal 1 ( l inear unit) . The length 1 ( l inear unit) would mean a closed interval

of len gth 1 ( l in ear unit) . St ruct ural l y this wo uld mea n that we can divide said closed

int erva l into two in terva ls, on e of wh ic h is a c lo sed inter val of leng th 1/2.

32. Al geb r a ica l l y w e know (A+B)2 = A

2 + AB + BA + В

2 = A

2 + 2AB + B

2. Geo-

metr ical ly, taking A = [OX], a c losed interval , and В = [OX), a onesided open inter-

va l ( t o b e re fer re d t o as an o p e n int e rva l ) , th e a b o v e a l g e b ra i c equ a l i t y w o u l d ac-

co un t for the inte rnal st ru cture of the squa re of leng th [О Х] .

Structura l BreakUp of a Squ are

33. Geomet r ica l l y t h e c l osed int erva l [OO'] can b e decomposed as t h e c l osed in-

terval [ OO '] = [O X ] + (X O'] = [O X ] + [OX ), wh ere X is the mi dd le poi nt of the

clo sed inte rval [OO '] . Th e structural bre ak up of a squa re of leng th 1 ( l ine ar unit )

can be in te rm s of the st ruc tural bre ak up perm iss ible by the l inear unit as a c los ed

int er va l [OX) of 1/2 unit length and open interval [OX) of 1/2 unit length. The

ab ov e f igure sho ws the struc tural brea k up of a square of leng th [OO '] wit h X as i t s

middl e point . Th e a l geb ra ic equ a l i t y [OO']2 = [ O X ]

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

2 w h en

translated into geometr ic equal i ty, as has been shown in the above f igure, would 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 in to four g e o m e t r i c a l s q u a re s . O u t o f t h e s e four s q u a re s ,

t h e f i r s t squ a r e represent ed b y [OX]2 i s a complete square with an area equal to the

squ are of the lengt h [O X ], a pe r im ete r equa l to 4 t im es the leng th of [O X ], and thefour boundary l ines and four corner points intact . The second and third squares rep-

resent ed b y 2[OX][OX) a re not compl et e squ ares , s ince one b ou ndary l ine and t w o

corner points are missing. The fourth square as wel l i s not complete, s ince i t s two

b o u n d a r y l i n e s an d t h re e c o r n e r p o i n t s are m i s s i n g .


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34. The original square may be reconst i tuted in the fol lowing four steps as depict

ed ab ov e. As a first step, out of the four squa re s take the squa re with four bo un da ry

line s and four co rne r poi nts intac t. As a se co nd step take one mor e squar e out of the

remaining three squares with only one boundary l ine and two corner points missing.

When this square is added to the square of the first step, it constitutes a rectangle of

are a equa l to half the area of the orig ina l squ ar e.

35. Here i t may be relevant to note that geometrical ly there remains no vacuum at

all when the two squares of equal length, one of them with its one boundary line

mis sin g, are join ed. This is be cau se [OX ] + [OX) = [OX ] + (XO' ] = [O O] , thus pro

viding a continuum throughout the boundary l ine along which the two squares are

j o i n ed to const i tu te a r ec tangle .

36. Now as a third step, we may take one more square whose one boundary l ine is

mis sin g from the rema ini ng two squar es. W he n this squa re is join ed with the recta n

gle composed in the first two steps, the missing boundary line, as is shown in the

abo ve figure, be co me s a co nti nuu m in term s of the half bou nda ry l ine of the inner

length of the rectangle.

37. For a fourth step, join the remaining fourth square, whose two boundary l ines

are missing, with the geometrical figure formed as a result of the first three steps. As

is evide nt from the above figure, on e of the miss ing bound ary l ines would be co me aco nt in uu m in ter ms of the upp er half of the boun da ry line of the inne r length of the

rec tan gle left unc ove red unti l Step 3, whi le the secon d missin g bou nda ry l ine of this

fourth square (as depicted as Step 4 in the above figure) would become a continuum

in terms of the inner boundary line of the third square.

38 . Thi s internal structural arra ngem ent of the square is significant in several ways .

Two of these which have vital bearing for the present are that the square has nine struc

tural poin ts out of wh ic h eight are symm etr ica lly loca ted aroun d the centr al ninth point

where all the four squares are joined, and that when out of the square of rational length,

a squa re of ratio nal len gth is cut out, the rem ai nin g porti on of the original sq uar e con

sists o f thre e parts , non e of wh ic h has all the four bou nda ry line s intact.

Internal Structure of a Cube

39. The cube is a three-dimensional regular body with the same length on al l the

three-dimensional l ines. Taking the length to be a uni ts, the cube has volume a3

and


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MOD ERN SCIENCE AND VEDIC SCIENCE

surface area 6a2. By sui tabl y cho os in g a l ine ar unit, we can hav e the len gth of the

cube = 1 (l inear unit). The length 1 (l inear unit) would mean a closed interval of

length 1 ( l inear unit) . Structural ly this would mean that we can divide the said closedint erv al into two inte rval s, on e of wh ic h is a clo sed inte rval of len gth 1/2.

40. Al geb ra ica l l y w e know t h at (A+B)3 = A

3 + 3 A

2B + ЗАВ

2 + В

3. Geomet r ica l l y

taking A = [OX], a c losed interval , and В = [OX), a onesided open interval ( to be

refe rred to as an op en interv al) , the abov e a lgebr aic equal i ty wo uld acc oun t for the

int er n a l s t r u ct u re of t h e cu b e of l engt h 2[ OX ].

Struc tural Brea kU p of a Cub e

4 1 . Geom et r ica l l y w e h ave [OO '] = [OX ] + (X O'] = [OX] + [O X ), w h e re X i s t h e

mi dd le po int of the close d interval [O O' ]. The structu ral break up of a squ are of

len gth 1 ( l ine ar unit ) can be had in ter ms of the st ructural break up perm iss ible by

the l inear unit as a c losed interval [OX] of 1/2 unit length and an open interval

l engt h [OX) of 1/2 unit leng th. The abo ve f igure sh ow s the structura l bre ak up of a

cu be of leng th [O O'] with X as i t s mi dd le poi nt . The algebr aic equal i ty [O O']3 =

[ O X ]3 + 3 [ O X ]

2 [ O X ) + 3 [ O X ] [ O X )

2 + [OX)

3, w h en t rans l a t ed int o a geomet r ic

equal i ty, as has been shown in the above f igure, divides the geometr ical cube into

e i g h t g e o m e t r i c a l c u b e s .

42 . Out of the abo ve eight cubes , the f irst cub e, repr ese nte d by [OX ]3

, is a com-

ple te cub e havi ng a vo lu me equal to a cu be of the length [ OX ], surface area equa l to

six t imes the area [OX]2

, and all the six surfaces and eight corner points intact.


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46. The division of the cube into eight cubes of above description can be used in re

ve rs e to rec ons titu te the ori gin al cu be in term s of the said eight cub es. Th is is simil ar

to the case of the squ are trac ed abo ve in pa ra gr ap hs 34 to 37. The said eight cube s


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have in all 36 surfaces. When these are combined as above, twelve surfaces are used to

form a conti nuu m with the missing surfaces of said cube s. Th e rema inin g 24 surfaces

account for the surface area of the re-composed original cube as is evident from the

above figure.

Summary

47 . Befo re we take up the gene ral Nth term of the framed dom ai ns sequ enc e, I would

like to sum ma ri ze the positi on eme rg ing from the abo ve analy sis of the first three

framed domains. These three cases involve common geometrical figures, namely a

clo sed inte rval of a straight line, a squa re, and a cub e of rationa l length s.

48. With respect to the first framed domain, it requires a definition of a structural

frame for a closed interval. With respect to the square and the cube it is obvious that

the are a of the squ are is cont ain ed within the perim ete r of the squa re and the vo lu me of

the cube is similarly contained within the surface area of the cube. As such the

pe rim et er and surfa ce area respe cti vel y are the fram ed par ts of the squar e and the cube.

Structurally the length of the closed interval requires a middle point and as such we

may define a point as a zero space figure responsible for providing a frame for the

(linear) length by having placement for the frame point anywhere in between the

interval. (To be specific, the placement for the frame point is not at the beginning or

the end point of the closed interval.) That way only one point (to be called the middle

poi nt since it falls mi dw ay bet we en the tw o end poin ts) const itut es the fram e of the

first framed domain.

49. From the above, we may conclude (and tabulate as the figure below indicates)

that when from the first framed domain, that is, a closed interval of a straight line, a

closed interval of smaller rational length is cut out, the remaining portion constitutes

an irrational length as it is missing one end point. When from the second framed

do ma in , that is, a squar e of ratio nal leng th, a squ are of sma ller rationa l len gth is cut

out, the rem ain ing portion constitutes three squ ares out of whi ch two square s are

missing one boundary line, and one square is missing two boundary lines. Similarly,

when from the third framed domain, that is, a cube of rational length, a cube of smaller

rationa l length is cut out, the rema inin g portio n constitutes sev en cubes out of whi chthree cubes are missing one surface, another three cubes are missing two surfaces, and

the last, that is, the seventh cube, is missing three surfaces.

50 . Th e follo wing table evid entl y mak es it clear that wh en the rational leng th of the

reg ula r bod y of the Nth- dim en si on (Nth frame d do ma in) is divi ded into rational part (Q)

and the remaining irrational part (R), the regular body gets divided into 2 N N-

dimensional bodies of the form Qn

RN - n

, wher e n = N, N- l , N- 2 , ... , 2, 1, 0. The se N-

dime nsio nal bodi es, numb eri ng N + l , may be called respective ly, first, second , . . .

(N+l)th-dimensional blocks of degree N. There is only one N-dimensional body of the

first-di mensiona l block. The re are N N-di mens iona l bodie s of the sec ond-di mens ional

b l ock . Th e R - d im en s io n a l b l ock h as ( N . N - l . N - 2 . . . N - R / l . 2 . 3 . . . R ) N - d im en s io n a l

bodies. The last block has one N-dimensional body.

5 1 . As a net result, the fol lowi ng internal a rra nge men ts of regul ar bod ies justify our

log ic and con clu sio n of pa ra gr aph s 7 to 12 that wh en from the rational length , a ration al

portion is cut out, it leaves behind an irrational length.


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Degree of Freedom

52 . No w we ma y take up the que stio n wh y restr ictio ns are nece ssa ry for n to be 3.

In other words, the question is: why for the first and second degree of natural numbers,

that is, for n = 1 and 2, the equation Zn

= Xn

+ Yn

has solutions. Here it may be relevant

to note that the division into two parts of like degrees is not possible for all Z n, n = 1, 2.

When Z and n both are equal to one, the division for Z n = 1 1 is not possible, though for

all Z 2 and n = 1, the des ire d div isi on int o tw o par ts obv iou sl y ho lds by the ve ry defi

nition of the natural num ber s. For exa mpl e, consi der 1, 1+ 1, 1+ 2, 1+3 .. ., 1+s ,. .. .53. For n = 2, every Z n is not decomposable as X n + Y n . Illustrations of the point are

l 2 , 2 2 , 3 2 , 4 2 . There are infinitely many values of Z for which the desired decomposition

is not availa ble. O n the con trar y, there are infinitely ma ny valu es of Z for wh ic h the de

sire d de co mpo si tio n holds . On e class of such valu es is Z = 5. The deco mpo sit ion for

this value is 5 2 = 3 2 + 4 2 . No w for any val ue of the form 5Z, a similar rela tions hip also

holds, namely (5Z) 2 = (3Z) 2 + (4Z) 2 .

54. So for n = 1 and 2, the desired decomposition is not universally permissible

tho ugh the sa me hold s for infinitely man y value s of Z. This restricte d perm issib ility of

the decompositions for n = 1 holds for all values except Z = 1. And for n = 2 for a large

number (a countable number of values of Z) the decomposition holds and for an equally

large num be r (a count able nu mbe r of valu es of Z), the sa me does not hold.

55. The reason for the restricted permissibility for n = 1 is that the absolute value of

natu ral num be r 1 and the linear mea su re value of unit lengt h are two distinct conc ept s.

The natural number 1 is the absolute value in the sense that it is not dependent upon any


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dimensional structures while the linear measure value of unit length is completely de

pendent upon the structural arrangement of a one-dimensional space, that is a straight

line . The line ar unit wo ul d repr ese nt a clos ed interva l of unit length. The inte rnal struc

tural restrictions of a closed interval make it impossible for its division into two closed

inte rva ls. As such the natura l num be r 1, as the first deg ree value of one , does not admit

the desired decomposition. However, the straight line and hence a closed interval, being

a first-dime nsional bod y (figure) has two degr ees of freedom within a three-dimen sional

space. As such, externally the linear unit is free to combine itself with a similar unit to

cons titu te a one-d ime nsi ona l bod y (figure) of two units. T he role of the origin point of

the three-dimensional frame is crucial since from it emanates dimensional lines and for

each of the three-dimensional lines it remains a zero value starting point. Because of

this, it is possible to have linear units simultaneously sprouting from the origin point

along the dimensional lines. Hence, there are two distinct planes available for every di

mensional line, which may be used by the given linear unit of any of the three dimen

sional lines. This two-fold freedom for the straight line accounts for the desired decom

posi t ion for all natural numbe rs except Z - l .

56. Turning to the plane, a square, a two-dimensional regular body, has as well, one

degr ee of free dom in a thre e-di mens iona l spa ce. It is bec aus e of the restricted freedom

of the plane that the deco mpos itio n is not universa lly permissi ble in the case of two -

dimensional regular bodies. Rather, in their case, within any finite range of natural

num be rs, say the desired dec ompo siti on general ly wou ld not be permi ssibl e

and in a com par ati vely very small nu mbe r of value s of Z only the desired decom pos i

tion would hold. This is so also within a three-dimensional space, two dimensions stand

restricted because of the structural format of a plane and it is left with only one degree

of freedom. Here we may compare the situation with the fate of a one-dimensional

body, that is, a straight line whose structural format has only one dimension. It is left

with two degrees of freedom in terms of unrestricted two dimensions of the three-

dimensional space.

57. Now when we come to a three-dimensional body, its structural format restricts all

the thre e dime nsi ona l lines and henc e we are left with no degr ee of freed om. It is be

ca us e of this that it is not possi ble to dupl ica te the cube . In the ca se of a pla ne , it is pos si

ble to duplicate it as a plane having one degree of freedom in a three-dimensional space.

As such we can move (or pile) a plane (or identical regular bodies of a two-dimensional

space, say square) along the third dimension as is evident from the following figure:


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58 . The first bas ic structural char acte rist ic acquire d by the origin point beca use of the

poss ibil ity for the free moti on of a plane of OX , OY frame alo ng OZ dim ens ion al line is

that the origin point gets embedded with a plane structure. Because of this, it hardlyma tte rs whi ch point of the Z axis is chose n as the origin point. T ho ug h the pla ne struc

ture acquired by the origin point of the dimensional frame does not affect the external

freedom for motion of the plane towards the OZ axis, the internal structural arrange

ment of the plane figures is governed by the two-dimensional format. The glaring effect

of the two-dimensional format for the plane figures is that its every point acquires a

two -di me nsi ona l structure. Beca use of this even the dimensio nal lines, as bou nda ry

line s of pla ne figures, get inse par ably me rge d with the plan e figure. As suc h, struct ural

ly they too can be dealt with only as plane strips as shown in the figures below.


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59. The immediate effect is that when we divide a square N2

, or as a matter of fact

even a rect angl e (N xM ), of lengt h N, then l2

units of the X axis, as has been shown

above, would be N, but the l2

uni ts of Y axis woul d remai n only N - l (in the case of a

rectangle, only M-l). Therefore, the re-assembled position for l 2 = 1 for the X and Y

axis would give N x ( N - l ) (and in case of a rectangl e N x [ M - l ] ) which is less than N2

(in the case of a rec tan gle N x ( M - l ) < N x M). Th is is the reaso n wh y in a large num be r

of cases it is not possi ble to deco mpo se Z 2 as X 2 + Y 2 for natural numbers Z, X, and Y.

60. Now usin g one degree of freed om, wh ich may be viewed as the external freed om

for the plane f igures, the restr ictions of the two- dim ensi onal format of the plane figures

can be kept in ab ey an ce . Her e it ma y be releva nt to note that the proof of the Py th ago ras

Th eo re m, wi th the help of a right-tr iangle, ma y appear as if we are using only one plane.

As a matter of fact, we are using more than one plane by considering the sides of a triangle. Suppose the square of the hypotenuse of a right-angle triangle is split into two

squares which equal the squares of the perpendicular and base of the said triangle or, as

a conver se, the rec onstr uctio n of a square of the hypo tenu se in term s of the squar es of

the base and perpendicular. Then we have to use the internal structural arrangments per

mitt ed by the squar e as a regular body of the two-dim ensi onal sp ace. Furthe r, we have

to use eight sy mm et ri es of the squ are and one degr ee of free dom (wh ich the squar e as a

two-dimensional body would have) within a three-dimensional Cartesian frame.

6 1 . As is sho wn abov e, on the two-d ime nsi onal format of OX , OY dime nsio nal lines

we can have plane figure ABCD (which may be a square or a rectangle). This figure

would have a degree of freedom of motion along the OZ axis. This provides us with the

possibility to choose two identical figures, say ABCD and A'B'C'D' as shown above.

The diagonal, being on the one-dimensional format, would have length equal to any realnumber, which includes the natural numbers. The two-dimensional figure AC C'A' may

be a square on a proper choice of vertices.

62 . It is be ca us e of the ab ov e freed om of con str uct ion for the plan e figures in a three-

dimensional frame that it becomes possible to divide the square of the hypotenuse (AC


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C'A' ) in term s of the squ are s on the dim en sio na l lin es of the forma t of the pla ne figure

ABCD and vice-versa. Though this assignment as such is not taken up here, it may not

be out of con tex t to note that the unit squar e wo ul d be the basi c con stit uen t wh ic hwo ul d be used for a divisi on proce ss of one squa re into two squar es or the reverse pro

cess of composition of two squares into one square. The unit square (and as such any

square) has eight symmetries, and the three-dimensional regular body as well has eight

corner points, and the three-dimensional Cartesian frame cuts the space into eight oc

tants, as shown below. It is because of this that the division process and the reverse

composition process is possible.

63 . He nc e we have justified the restrictio n of Fe rm at 's Last Theor em to n ^ 3.

64. Now we may 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 bodies

moving from two-dimensional bodies to three- and higher-dimensional bodies.

Case o f a Three - Dim ens i ona l Reg ul a r Body

65. Step 1: Let volume V = Z 3 cubic units. In other words, the cube Z is constituted

by Z 3 unit cubes.

66. Step 2: The cube Z has Z units along the OX dimensional line as well as Z units

along both of the remaining two dimensional lines.


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68. Step 4: Z unit cubes are required to constitute a length of Z units along the Y axis

and similarly, Z unit cubes are required to constitute a length of Z units along the Z axis.

69. Ste p 5: The ori gin point of the Cartesian frame, bein g a dime nsio nles s point, wh en

the X axis and the Y axis emerge 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.


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72. Step 8: We had started with cube Z of length Z and volume V = Z3

, but have

end ed with a cub e of vol ume Z (Z+ 1)( Z+1 ) whi ch is greater than Z 3 , hence there is a

contradict ion.

Conclusion

73 . Ther efor e, the abov e contra diction prove s that our ass umpti on rega rding permis

sibility of the vo lu me of the cube to be sepa rate d and han dle d l x l x l = 1 as a natur al

unit is wrong. Hence X3

cubic units cannot be treated as linear units and as such the ad

dition operatio n of natural num ber s is not a meanin gful operation. If cubic units l x l x l

are treated as linear units 1, then it would result in a contradiction. Hence, the theorem

for the cubes that it is impossible to separate a cube into two cubes.

74. The above logic would be equally applicable to the bi-quadratic and higher pow

ers as well, as the natural numbers addition operation is geometrically linear in nature;

the same is mea ning les s and it is not applic able to the higher-di mens iona l units l x l ,

l x l x l , l x l x l x l . . . .


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MOD ERN SCIENCE AND VEDIC SCIENCE

75. Z2 ( l x l ) a n d Z

2 (1) are two dist inct units . Similar ly Z

3 ( l x l x l ) a n d Z

3 (1) as well

a s Z4 ( l x l x l x l ) a nd Z

4 (1) and so on, are pairs of dist inct units . The a ddit io n oper at io 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 fo r th e u n i t s l x l o r l x l x l , o r l x l x l x l a nd s o o n , s in ce( l x l ) + ( l x l ) mea n s an op er at ion in t e r ms o f w hic h t w o squar e un i t s ar e t o be com-

b i n e d , w h i l e 1 + 1 = 2 is t h e n a tu r a l n u m b e r s ad di t ion o p e r a t io n . U n l e s s and unti l w e

know how the square unit is t ransformable into a l inear unit , the l inear units addit ion op-

er at ion w oul d n ot be ap p l ic abl e . The n at ur a l n umber s mul t ip l ic at ion op er at ion , w hic h i s

in t imat e l y c on n ec t ed w i t h t he addi t ion op er at ion , r emain s dep en den t up on t he addi t ion

operat ion only within a l inear space. The moment the space stands changed from one

dimen s ion al ( l in ear ) sp ac e t o t w o an d h igher d imen s ion al sp ac e , t he in dep en den t c har -

acter i st ics of the mult i pl ica t ion opera t ion are displayed . As such, unle ss and unt i l l x l i s

suita bly def ined, the addit i on opera t ion wil l rem ain only the ope rat io n of one

d i m e n s i o n a l s p a c e .

Geometrical Continuum: FourthDimensional Space

7 6 . N o w w e may t ak e up t he four t h n um be r o f t he f r am ed do ma in s se qu en c e

[an/2na

n1 fo r n = 1, 2, 3 ... ]. F or the pre sen t it ma y be taken by wa y of defi nitio n that

[a4/8a

3] i s the formul at io n for the regular bod y of four thdi men siona l space of dimen-

s ion al l en gt h a, c on t en t p ar t ( domain p ar t ) as a4, and the frame part as 8a

3. In the con-

text we may refer to a cu be of dim ens ion al leng th a , c on t en t ( domain p ar t ) as v o l ume a3,

and frame part as sur face area 6a2 as the threedimensional regular body. Further, by

w a y o f d e f ini t ion , w e m a y ta k e t h e fo l l ow in g figure as a g e o m e t r i c a l p r e se n t a t io n o f t he

four t hd imen s ion al r egul ar body .

Structural Arrangem ent of Fourth Fra med Dom ain

7 7 . Al gebr a ic a l l y w e k n ow t hat ( A+ B )4 = A

4 + 6 A

3B + 4A

2B

2 + 6AB

3 + B

4. Geomet-

rically, taking A = [OX] and В = [OX) as a onesided open interval (to be referred to as

an op en interval) , in ter ms of the above expa nsi on, A+ B = [OO'] = [OX ] + (X O'] =

[OX] + [OX), when X is a rat ional number and is the middle point of [OO']. I f [OX] =

X , t h e n [ O O ' ] = 2 X .

7 8 . T h e a l g e b r a i c e q u a l i t y [ O O ' ]4 = [ O X ]

4 + 6 [ O X ]

3 [ O X ) + 4 [ O X ]

2 [ O X )

2 +

6 [ O X ] [ O X )3 w o r k s out t he s t r uc t ur a l ar r an gem en t o f t he four t h f r amed do ma in . Thi 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 th e f o u r t h d i m e n s i o n a l b o d y in t o se ve 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 .

7 9 . Out o f t he sev en t ee n four t hd imen s ion al bod ies , t he fi rs t, n amel y [OX ]4

, is a

c omp l e t e four t hd imen s ion al body in t he sen se t hat a l l t he e ight c ubic c omp on en t s o f

the frame are intact .


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80. We may define a body of such frame with all the eight cubic components intact

as a first-dimensional block of the regular body.

8 1 . As is evident from the above a lgeb raic equa lity, in all we shall have five types of

dimensional blocks. The second, third, fourth, and fifth-dimensional blocks would be

structurally the fourth-dimensional bodies such that their frames are respectively miss

ing 1, 2, 3, and 4 cubic components.

82. Theref ore, the internal structural arrang eme nt of the fourth-dime nsional regula r

body admits a break-up as five-dimensional blocks such that these blocks have respec

tively 1, 6, 4, 6, and 1 fourth-dimensional bodies. Evidently, all the five-dimensional

blocks are structurally distinct.

83 . No w the general binomial expans ion of (A +B )N

wou ld hel p us con clu de that we

shall have (N+l)-dimensional blocks. In this format, 2 N N-dimensional bodies would

constitute the regular body (A+B) N . As for the N-dimensional space, the format of the

dimensional blocks is to remain the same, so as to structure a regular body, exactly 2N

N-dimensional bodies would be required.

84 . Th er ef or e, for N 3 and Z 3 (for Z = 1 and 2 the th eo re m is ob vi ou s) , let ZN =

XN

+ YN

where X, Y, Z, and N are natural numbers. Further, let Z > X > Y (for X = Y

the theorem has been proved above, and X and Y being general, it remains to prove the

the or em wh en one of X or Y is grea ter than the other). L et Z = X + R. The ref ore, X N +

Y N = (X+R) N . The right side on expansion yields X N + NR X N - 1 + .... Therefore, X N + Y N

= XN + NR X N - 1 + ... + R N .


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Therefore, Y N = N R X N - 1 + ... + R N .

No w the right side is an exp ressi on of 2 N -1 N-dimensional blocks.

The left side can have divisions into N-dimensional blocks as Y = A+B.

Y N = (A+B) N yields 2 N N-dimensional blocks, while the N-dimensional blocks on the

right side are only 2 N - 1 . Hen ce , the left side canno t be equal to the the right side. Wi th

this the geomet rical pro of of the theo rem stan ds comp leted .

85. Now if in the light of the above, we try to find the significance of the internal

structu ral arr ang emen t of the very first regular body l1

, where 1 is equal to an irrational

length, then it really would be enchanting to see how Nature works out 1 = 1/2 (half of a

rational unit) + 1/2 (half of an irrational unit). It provides us with a geometrical continu

um whic h, as a linear con tin uum , is equivalent to the field of real numb ers who se sub -

field is the field of rational numbers.

86. Obviously, the implications are many and far reaching, particularly when they

demo lish m enta l blocks of the physical world and help transcend to four and higher-

dimensional spaces as a geometrical continuum of spaces whose regular bodies consti

tute a framed domains sequence (a n / 2 n a n - 1 , n = 1, 2, 3 ...).

Documents

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