Vedic Mathematical Concepts and Their Application to ... Mathematical Concepts and Their Application...

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

Indian Institute of Maharishi Vedic

Science and Technology

Preface

The following three proofs, based on the application of Vedic Mathemat ica l con

cepts, address a famous unsolved problem of mathemat ics , Fe rmat ' s Last Theorem. The

fol lowing passage of Jabali Upanishad provided the structural key for developing the

mult i-dimensional spaces used in the argument for Fe rmat ' s Last Theorem:

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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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 earlier work, The Rationale of Оm—Its Formulation, Significance and Occur-

rence and Applications in Ancient Literature, I commented on the above passage 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.

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 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 ign i f i cance of the above wi th in the a r i thmet ic d o m a i n is that in o rder to

u n d e r s t a n d the s t ruc tu ra l f rames and s y s t e m s of natural 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 ) we have to d e v e l o p the unde r s t and ing in t e rms of the

s t ruc tu ra l f rames and s y s t e m s of na tura l n u m b e r 6 (in U p a n i s h a d i c l anguage S h a d a

nan 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 Jaba l i R ish i ) . A n d u n d e r s t an d i n g the s t ruc tura l

f r ames and s y s t e m s of na tura l n u m b e r 6 mus t be in t e rms of the s t ruc tura l f rames

and s y s t e m s of na tura l n u m b e r 4 (in the U p a n i s h a d i c l anguage Ishan b e s t o w e d en

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 eve r se sequen t ia l p rocess con ta ins the s t ruc tu

ral key .

T h e bas ic V e d i c Mathemat ica l concep ts used are that the unity (s ingle-syl lable

O m ) is p rocessab le quar ter by quar ter (Shri Pada processing line) and the fourth

quar ter is the integrat ion of the first three quar ters (Maharishi processing line). As

such, I interpret the above Upan i shad ic passage in the fol lowing way: the structural

f rames and sys tems of natural number 4 (and hence the four th-dimensional domain)

are to be handled as uni ty , admit t ing process ing quarter by quarter, and the fourth

76

VEDIC MATHEMATICAL CONCEPTS

quar te r ( the unmanifes ted quarter) is to be processed as integrat ing the first three

quar te r s (manifes ted quar ters) . The process ing within the fourth quarter a long the

Nor th -Eas t d iagonal wou ld lead to the domain of Shadanan (admit t ing structural

f rames and sys tems of natural number 6 and hence a s ix-dimensional domain ) . The

U p a n i s h a d i c c o m m a n d is to reverse the process , as the process ing is to be had in the

d e s c e n d i n g order from 9 to 6 and 6 to 4, but here in this world everything is hang ing

ups ide d o w n u r d h v a m u l a m — B h a g a v a d - G i t a , 15.1). This means that

wh i l e our a im is to p rocess within the unmanifes ted quarter in order to go from the

four th-d imens iona l domain of natural n u m b e r 4 to the s ix th-dimensional domain of

natural n u m b e r 6, we have to process a long the Nor th-Eas t d iagonal but in the di rec

t ion w h i c h leads from the s ix th-d imens ional domain towards the four th-dimensional

d o m a i n . Fur ther , the process ing must be quarter by quarter, which means that w h e n

natural n u m b e r 4 is taken as a uni ty, its quarter would be one, and similar ly, w h e n

natural number 6 is taken as a unity, its process ing unit would be one . Hence the p ro

cess ing a long the Nor th-Eas t d iagonal is to be had by dividing i t into six par ts . Out of

these six par ts , only one part would be unmanifes ted and the remain ing five parts

w o u l d cover the manifes ted part . Wi th this the procedure of structural p rocess ing

s tands as fol lows:

Step 1: The processing is to be in the fourth quarter of the Om formulation.

Step 2: Fourth quarter (fourth component ) of the Devanagar i Om formulation is like

a two-dimensional cartesian frame.

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

cess ing on the format of a plane.

Step 4: As both the dimensional lines are symmetr ic , the natural figure of the plane

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

Step 5: For processing of the first three quarters of the manifested domains , divide

the square 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 processing

unit wou ld be Z /3 .

Step 7: As the fourth quarter is to be processed as the integration of the first three

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

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

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

Step 8: The processing square be ing of length Z, the manifested part of the North-

East d iagonal of the extended processing square would be 5Z/6, and the unmanifested

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 ,

w h i c h means we are t ranscending from the geometrical square to the values-square (as

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

metr ical square itself in terms of the cross points of the lines parallel to the axes of the

two-dimens ional cartesian frame of the geometr ical square. If we take into account only

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

then the cross points on the length or breadth of the square would be equal to the cross

poin ts on any of the two diagonals of the square .

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Step 10: The effect of extending the processing square into the extended processing

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

tended processing square is the structural key for resolving why the sum of two 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 Fe rma t ' 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 appeal ing to

other scholars to approach the main chal lenges within their disciplines through the Vedic

w i sdom.

T h e Vedic perspect ive on methodology integrates objectivity with subjective experi

ence and as such those w h o are trained in objective methodology are required only to

learn how to supplement this approach with their own subjective experience of the Vedic

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

baji w h o initiated me for the Shri Pada processing line to process this quarter by quarter,

and at the feet of His Hol iness Maharishi Mahesh Yogi w h o initiated me for the Mahar i -

shi process ing line to process the fourth quarter as the integration of the first three quar

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

harishi Vedic Science and Technology, for experienced guidance and personal interest

in these and other studies at the Institute.

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Proof by Direct Comparison

Overview


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80


81


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84

Statement of the Theorem


85

Therefore, T is not a rational number .

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

fore, we have to look for solutions of this equation in the field of real numbers . Here it

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

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

bers . This has the obvious solution in the field of rational numbers . However , when the

solution is required for U N = V N + T N , where V and T are irrational numbers , we have to

shift to the field of real numbers .

Step 6: The field of real numbers consti tutes the arithmetical cont inuum which is

equivalent to the linear geometr ical cont inuum of a straight line. N o w the solutions of

the equat ions U N = V N + T N and M N . U N = M N . V N + M N . T N are directly linked up. There

fore, the solution of the equation ( M U ) N = ( M V ) N + ( M T ) N , which is nothing but S N =


po in t r ep resen t ing the n u m b e r zero (0) . As a second s tep, for a g iven ra t ional n u m

ber Z, we may cut a c losed interval [OZ] of length OZ = Z from the l ine OR. As a

th i rd s tep , for any rat ional n u m b e r X < Z, we can cut a c losed interval [OX] of

l eng th OX = X from the interval [OZ] of length OZ = Z.

4 . T h e in terval [OZ] gets d ivided into two par ts , namely c losed interval [OX] and

the one - s ided open interval ( X Z ] . Bo th the par ts , name ly [OX] and ( X Z ] , do not

h a v e any c o m m o n point . All n u m b e r s ra t ional and irrat ional less than or equal to X

are in [ O X ] , and all n u m b e r s rat ional or i rrat ional greater than X are in ( X Z ] . As

such , the interval (XZ] is of an i r ra t ional length.

Fermat's Last Theorem 5. S ta tement : "It is imposs ib le to separa te a cube into two cubes , or a b iquadra te

in to two b iquadra tes , or in genera l any power higher than the second into p o w e r s of

l ike d e g r e e . " — Fermat

6. In modern mathematical language this is restated as: x n + y n = z n 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 n u m b e r s , therefore z n and x n are also natural n u m b e r s .

Let Z = z n and X = x n . Therefore , V = Z - X .

9. As Z and X are natural n u m b e r s , so the c losed interval [OZ] of length Z cuts a

ra t ional length on the real l ine OR. S imi lar ly , the c losed interval [OX] of length X

cu t s a ra t ional length from the interval [ O Z ] .

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

1 1 . H e n c e , V N" for every natura l n u m b e r N, as N n i s a natural n u m b e r so

w o u l d equal a ra t ional length and not an irrat ional length .

12. T h e above p roves the theo rem but for the restr ict ion for the n to be 3.

Restrictions for n to be 3 Rat iona le for the res t r ic t ions :

13 . To arr ive at the ra t ionale for the res t r ic t ions for n to be 3, we have to go

b a c k to our defini t ion of the d imens ion in t e rms of its geomet r i ca l represen ta t ion .

14. T h e defini t ion of d imens ion canno t be expressed except wi th in the mani fes t

ed w o r l d of th ree -d imens iona l ob jec ts , thereby pe rmi t t ing its a r r angemen t as a l ine

ar g e o m e t r i c a l c o n t i n u u m of s t ra ight l ines , each a ma themat i ca l mode l of the ari th

me t i ca l c o n t i n u u m of real n u m b e r s . In concre te t e rms , geomet ry accep ts d i m e n s i o n

as that w h i c h comple t e ly pe rmi t s represen ta t ion as a s t raight l ine, and the g e o m e t r i

cal un ive r se fo l lowing is the t h ree -d imens iona l space wi th the three d i m e n s i o n s

no th ing but s t ra ight l ines . In fact, here l ies the ra t ionale and the answer to the ques

t ion w h y res t r ic t ions are p laced upon 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

mani fes ted wor ld cannot be accep ted as a proof as such of the ex is tence or o ther

w i s e of the h igher - d imens iona l spaces . For this we have to have a pure ly m a t h e m a t

ical approach .

15. For th is , we may have insight in to the internal s t ructure of N n by accep t ing i t

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to be a regu lar body of length N of n th-d imens iona l space (for the present take as a

def ini t ion that degree n represent s the numbe r of d imens ions ) .

The First Dimensional Regular Body

16. To have proper insight into the internal s t ructure of h igher-d imens iona l regu

lar bod ie s , we wou ld be requi red to p in-point the rat ionale as to why the very first-

d imens iona l regu lar body ( ll) does not permi t div is ion into smal le r regular bod ie s o f

the f i r s t-d imens ional space .

17. The answer is becau se of the very def ini t ion of the natura l numbe r 1. The nat

ural numbe r 1 is the smal les t na tura l number , so the ques t ion of the ex i s tence of still

sma l l e r natura l numbe r s doe s not ar ise. Hence l1 X

1 + Y

1, whe re X and Y are natu

ral numbe r s .

18. The above s i tuat ion deserves p rope r ma themat i za t ion . First ly, i t takes one as

unde f ined . S imu l t aneous l y i t g ives us the f reedom to accept any l inear length, may it

be equa l to ra t iona l or i r rat ional un i t s/numbers , as a l inear unit and hence " o n e . "

19. Us ing the above f reedom of cho ice to accept any l inear length as the dimen

siona l unit he lp s us to r educe any regular body N" to the first regular body of nth-

d imens iona l space ( Г ) , whe r e N = one uni t . Th i s also can be taken as the first regu

lar body of the f i r s t-d imens ional space ( l1) s ince 1

n = 1 = l

1.

20. Therefore, in order to understand the internal structural arrangements of dimen

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

for transforming unit l inear length into a rational fraction plus an irrational fraction.

2 1 . Des i gna te the point of s tudy as a structural knot of dimensional regular

bodies. We focus on the s t a tement that one l inear unit is equal to a half-unit ra t ional

l eng th p lu s a half-unit i r ra t ional length.

22 . Before we take up the above s t a tement for e laborat ion, I wou ld l ike to add

here that the ra t iona le for the above lies in the first pr inc ip le (of mani fes ted wor ld

admi t t i ng s t ruc tures ) that the s t ructura l uni t is half as compa red to one as a dimen

siona l uni t .

Definition of a Straight Line

23 . One at t r ibute of the straight l ine is that it has a midd le point . A second attrib

ute of the s tra ight l ine is that it has a m i n imum of three po in t s . It may not be possi

ble to g ive a prec i se def ini t ion of a s t ra ight l ine in te rms of s ome at t r ibutes . How

ever, the basic attributes of the straight line would help us settle some practical defini

t ions . The pract ica l def in i t ions may have wor th for any doma in of pract ica l impor

tance but the s ame may not be accep tab le to ma thema t i c s . A prec i se mathemat i ca l

def in i t ion of a s t ra ight l ine can be p rov ided by the ar i thmet ica l con t inuum of the real

numbe r s . The format benea th the ordered display of the set of real numbe r s wou ld

ma thema t i c a l l y qualify to be des igna ted as a straight l ine. As such this is accepted as

the def in i t ion of the s tra ight l ine for the purpose of expres s ions of d imens iona l

f rames , as we l l as for the purpose of cu t t ing lengths from a regular body .

Structural Unit Versus Dimensional Unit

24. N o w let us take up the cha l l enge of the s t ructura l knot (paragraph 21) . Here

the s i tua t ion p roceed s in two s teps : firstly, in te rms of the answer to the poser, i f we

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can t r anscend the three d imens iona l geomet r i ca l un iverse and have a geomet r i ca l

c o n t i n u u m , and secondly , in t e rms of the fine connec t ion be tween the s t ructural unit

and the d imens iona l unit ma in ta in ing the rat io of 1/2 : 1.

Square and Cube 2 5 . For both the s teps , init ial ly we m a y proceed wi th the s tructural c o m m o n n e s s

of the square as a regular body of a two-d imens iona l space and the cube as a regular

b o d y of a th ree -d imens iona l space .

26 . T h e s igni f icance of the s t ructural c o m m o n n e s s of a square and a cube is that

the ra t io of area and pe r ime te r of the square , and v o l u m e and surface area of the

c u b e , admi t sequent ia l o rder as a function of the degree of the d imens iona l space .

We may def ine the degree of n -d imens iona l space as n . T h e s e rat ios for length 'a' of

the regular bod ie s is:

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

2 7 . T h e above ra t ios have the formula t ion 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 a n (which for the square is the area of the square and for the cube is

its vo lume) 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 l 1 , l 2 , 1 3 . . . 1 n .

Framed Domains Sequence 2 9 . Before we inves t iga te the in ternal s t ructure of regular bod ies of unit length of

different d imens iona l spaces , let us unders t and the sequence [an/2nan-1 for n =

1,2,3...]. Th i s s equence may be def ined as a f ramed d o m a i n s sequence and its indi

v idua l t e rms as ind iv idua l f ramed d o m a i n s or s imply framed doma ins . T h e reason

for the cho ice of t e rmino logy is that the regular bod ies have their doma in part con

ta ined wi th in the frame part and thus can be des igna ted as framed doma ins .

First Framed Domain 3 0 . T h e first m e m b e r of the f ramed d o m a i n s 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 f ramed doma in .

T h i s ind ica tes that out of any c losed l inear interval we can r e m o v e one c losed inter

va l of half length wh i l e the second half wou ld not be a c losed interval . H e n c e the

s t ruc tura l unit is half of the d imens iona l unit , wh ich essent ia l ly is the l inear unit .

O n e m a y rever t back to this f ramed d o m a i n after hav ing been through the internal

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s t ruc ture of second and third framed doma in s , wh ich are the we l l -known geometri

cal f igures of a square and a cube , to fully real ize this impl ica t ion.

Internal Structure of a Square

3 1 . The square i s a two-d imens iona l regular body wi th the same length on both

d imens iona l l ines . Tak ing length to be a uni t s , the square wou ld have an area a2 and

a pe r ime te r 4a. By sui tably choos ing a l inear unit, we can have the length of a

squa re equa l 1 ( l inear un i t ) . The length 1 ( l inear unit) wou ld mean a c losed interval

of l eng th 1 ( l inear uni t ) . St ructura l ly this wou ld mean that we can div ide said c losed

interva l into two intervals , one of wh i ch is a c losed interval of length 1/2.

32 . A lgebra ica l l y we know (A+B)2 = A

2 + AB + BA + В

2 = A

2 + 2AB + B

2. Geo

metr ica l l y , t ak ing A = [OX] , a c losed interval , and В = [OX), a one-s ided open inter

val ( to be referred to as an open interval) , the above a lgebra ic equal i ty wou ld ac

coun t for the internal s t ructure of the square of length [ОХ] .

Structural Break-Up of a Square

3 3 . Geomet r i ca l l y the c losed interval [OO'] can be decomposed as the c losed in

terval [OO' ] = [OX] + (XO' ] = [OX] + [OX) , where X is the midd le point of the

c losed interval [OO ' ] . The structural b reak-up of a square of length 1 ( l inear uni t )

can be in t e rms of the s t ructura l b reak-up permis s ib le by the l inear unit as a c losed

interva l [OX) of 1/2 uni t length and open interval [OX) of 1/2 unit length . The

above figure shows the s t ructura l b reak-up of a square of length [OO'] wi th X as its

m idd l e point . The a lgebra ic equal i ty [OO']2 = [OX]

2 + 2 [OX] [OX) + [OX)

2 when

t rans la ted into geomet r i c equal i ty , as has been shown in the above figure, wou ld di

vide the geomet r i ca l square into four geomet r i ca l squares . Out of these four squares ,

the first square r ep re sen ted by [OX]2 is a comp le t e square wi th an area equal to the

squa re of the length [OX] , a pe r imete r equal to 4 t imes the length of [OX] , and the

four bounda r y l ines and four corner po in t s intact . The second and third squares rep

re sen ted by 2 [OX ] [OX) are not comp l e t e squares , s ince one boundary l ine and two

co rne r po in t s are mis s ing . The fourth square as wel l i s not comple te , s ince its two

bounda r y l ines and three co rne r po int s are mis s ing .

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3 4 . T h e or ig ina l square m a y be recons t i tu ted in the fo l lowing four s teps as depic t

ed above . As a first s tep , out of the four squares take the square wi th four bounda ry

l ines and four corner po in ts intact . As a second s tep take one more square out of the

r e m a i n i n g three squares wi th only one bounda ry l ine and two corner poin ts miss ing .

W h e n this square is added to the square of the first s tep , i t cons t i tu tes a rec tangle of

area equal to half the area of the or ig inal square .

3 5 . Here i t may be re levant to note that geomet r ica l ly there r ema ins no v a c u u m at

all w h e n the two squares of equal length , one of them wi th its one bounda ry l ine

mi s s ing , are jo ined . Th i s is because [OX] + [OX) = [OX] + (XO' ] = [ O O ] , thus p ro

v i d i n g a c o n t i n u u m th roughou t the b o u n d a r y line a long wh ich the two squares are

j o i n e d to cons t i tu te a rec tangle .

3 6 . N o w as a third s tep , we may take one m o r e square w h o s e one b o u n d a r y l ine i s

m i s s i n g from the r ema in ing two squares . W h e n this square i s jo ined wi th the rec tan

g le c o m p o s e d in the first two s teps , the miss ing bounda ry l ine, as is s h o w n in the

above f igure , b e c o m e s a c o n t i n u u m in t e rms of the half bounda ry l ine of the inner

l eng th of the rec tangle .

3 7 . For a fourth s tep, jo in the r e m a i n i n g fourth square , w h o s e two bounda ry l ines

are mi s s ing , wi th the geomet r i ca l f igure formed as a result of the first three s teps . As

is ev iden t from the above f igure, one of the miss ing bounda ry l ines wou ld b e c o m e a

c o n t i n u u m in t e rms of the upper half of the bounda ry line of the inner length of the

r ec tang le left uncove red unti l S tep 3 , w h i l e the second miss ing bounda ry l ine of this

four th square (as depic ted as S tep 4 in the above f igure) wou ld b e c o m e a c o n t i n u u m

in t e r m s of the inner bounda ry l ine of the third square .

38 . This internal structural arrangement of the square is significant in several ways .

T w o of these which have vital bearing for the present are that the square has nine struc

tural points out of which eight are symmetrical ly located around the central ninth point

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

a square of rational length is cut out, the remaining portion of the original square con

sists of three parts , none of which has all the four boundary lines intact.

Internal Structure of a Cube 3 9 . T h e cube is a th ree -d imens iona l regular body wi th the s ame length on all the

th ree -d imens iona l l ines. Tak ing the length to be a uni ts , the cube has v o l u m e a 3 and

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surface area 6a2. By suitably choos ing a l inear unit, we can have the length of the

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

length 1 (l inear uni t) . Structural ly this wou ld mean that we can divide the said closed

interval into two intervals, one of which is a closed interval of length 1/2.

40 . Algebra ica l ly we know that (A+B)3 = A

3 + 3A

2B + ЗАВ

2 + В

3. Geomet r ica l l y

t ak ing A = [OX] , a c losed interval , and В = [OX), a one-s ided open interval (to be

referred to as an open interva l) , the above a lgebra ic equal i ty wou ld account for the

in te rna l s t ructure of the cube of length 2 [OX] .

Structural Break-Up of a Cube

4 1 . Geomet r i ca l l y we have [OO'] = [OX] + (XO' ] = [OX] + [OX) , where X is the

m idd l e po int of the c losed interval [OO ' ] . The structural b reak-up of a square of

l eng th 1 ( l inear uni t ) can be had in te rms of the structural b reak-up permiss ib le by

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

l eng th [OX) of 1/2 unit length. The above figure shows the s tructural b reak-up of a

cube of length [OO'] wi th X as its midd le point . The a lgebra ic equal i ty [OO' ]3 =

[OX]3 + 3 [ O X ]

2 [OX) + 3 [OX] [OX)

2 + [OX)

3, when trans lated into a geomet r i c

equa l i ty , as has been shown in the above figure, d iv ides the geometr ica l cube into

e ight geomet r i ca l cube s .

42 . Out of the above eight cubes , the first cube , represented by [OX]3, is a com

plete cube hav ing a vo l ume equal to a cube of the length [OX] , surface area equal to

six t imes the area [OX]2, and all the six sur faces and eight corner point s intact .

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4 6 . T h e division of the cube into eight cubes of above descript ion can be used in re

verse to reconsti tute the original cube in terms of the said eight cubes . This is similar

to the case of the square traced above in paragraphs 34 to 37 . The said eight cubes

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have in all 36 surfaces. W h e n these are combined as above, twelve surfaces are used to

form a con t inuum with the missing surfaces of said cubes . The remaining 24 surfaces

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

above figure.

Summary 47 . Before we take up the general Nth term of the framed domains sequence, I would

like to summar ize the position emerging from the above analysis of the first three

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

c losed interval of a straight line, a square, and a cube of rational lengths.

4 8 . Wi th 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 area of the square is contained within the per imeter of the square and the vo lume of

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

per imeter and surface area respectively are the framed parts of the square 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

( l inear) length by having p lacement for the frame point anywhere in be tween the

interval . (To be specific, the p lacement 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

point since it falls midway be tween the two end points) consti tutes the frame of the

first framed domain .

4 9 . F rom the above, we may conclude (and tabulate as the figure be low indicates)

that w h e n from the first framed domain , that is, a closed interval of a straight l ine, a

c losed interval of smal ler rational length is cut out, the remaining port ion const i tutes

an irrational length as i t is miss ing one end point. W h e n from the second framed

domain , that is, a square of rational length, a square of smaller rational length is cut

out, the remain ing port ion consti tutes three squares out of which two squares are

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

w h e n from the third framed domain , that is, a cube of rational length, a cube of smaller

rat ional length is cut out, the remaining port ion const i tutes seven cubes out of which

three cubes are miss ing one surface, another three cubes are miss ing two surfaces, and

the last, that is, the seventh cube , is miss ing three surfaces.

50 . The following table evidently makes it clear that when the rational length of the

regular body of the Nth-dimension (Nth framed domain) is divided into rational part (Q)

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

dimensional bodies of the form Q n R N - n , where n = N, N - l , N - 2 , ... , 2, 1, 0. These N-

dimensional bodies , number ing N + l , may be called respectively, first, second, . . .

(N+l ) th -d imens iona l b locks of degree N. There is only one N-dimensional body of the

f irst-dimensional block. There are N N-dimensional bodies of the second-dimensional

b lock . The R-d imens iona l b lock has ( N . N - l . N - 2 . . . N - R / l . 2 . 3 . . . R ) N-d imens iona l

bodies . The last b lock has one N-dimensional body.

5 1 . As a net result, the following internal arrangements of regular bodies justify our

logic and conclusion of paragraphs 7 to 12 that when from the rational length, a rational

port ion is cut out, it leaves behind an irrational length.

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Degree of Freedom 52 . N o w we may take up the question why restrictions are necessary 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 Z n = X n + Y n 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.

W h e n 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 desired division into two parts obviously holds by the very defi

nition of the natural numbers . For example , consider 1, 1 + 1 , 1 + 2 , 1+3 . . . , 1+s , . . . .

5 3 . 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 decomposi t ion

is not available. On the contrary, there are infinitely many values of Z for which the de

sired decomposi t ion holds. One class of such values is Z = 5. The decomposi t ion for

this value is 5 2 = 3 2 + 4 2 . N o w for any value of the form 5Z, a similar relationship also

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

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

though the same holds for infinitely many values of Z. This restricted permissibility of

the decomposi t ions 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 decomposi t ion holds and for an equally

large number (a countable number of values of Z), the same does not hold.

5 5 . The reason for the restricted permissibili ty for n = 1 is that the absolute value of

natural number 1 and the linear measure value of unit length are two distinct concepts .

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

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

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

l ine. The linear unit would represent a closed interval of unit length. The internal struc

tural restrictions of a closed interval m a k e it impossible for its division into two closed

intervals . As such the natural number 1, as the first degree value of one, does not admit

the desired decomposi t ion . However , the straight line and hence a closed interval, being

a f irst-dimensional body (figure) has two degrees of freedom within a three-dimensional

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

const i tute a one-dimensional body (figure) of two units. The 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, i t is possible to have linear units s imultaneously sprouting from the origin point

a long the d imensional l ines. Hence , there are two distinct planes available for every di

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

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

posit ion for all natural numbers except Z - l .

56 . Turn ing to the plane, a square , a two-dimensional regular body, has as wel l , one

degree of f reedom in a three-dimensional space. It is because of the restricted freedom

of the p lane that the decomposi t ion is not universally permissible in the case of two-

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

numbers , say the desired decomposi t ion generally would not be permissible

and in a compara t ive ly very small number of values of Z only the desired decompos i

tion wou ld hold. This is so also within a three-dimensional space, two d imens ions stand

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

of f reedom. 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

wi th two degrees of freedom in terms of unrestricted two d imens ions of the three-

d imens iona l space .

57 . N o w w h e n we come 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 be

cause of this that it is not possible to duplicate the cube. In the case of a plane, it is possi

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

As such we can m o v e (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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5 8 . The first bas ic structural characteristic acquired by the origin point because of the

possibil i ty for the free motion of a plane of OX, OY frame along OZ dimensional line is

that the origin point gets embedded with a plane structure. Because of this, it hardly

mat ters which point of the Z axis is chosen as the origin point. Though the plane struc

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

f reedom 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-d imens iona l structure. Because of this even the dimensional lines, as boundary

l ines of plane figures, get inseparably merged with the plane figure. As such, structural

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

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59. The immediate effect is that when we divide a square N 2 , or as a matter of fact

even a rectangle (NxM) , of length N, then l 2 units of the X axis, as has been shown

above, would be N, but the l 2 units of Y axis would remain 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 rectangle N x [ M - l ] ) which is less than N 2

(in the case of a rectangle N x ( M - l ) < N x M) . This is the reason why in a large number

of cases i t is not possible to decompose Z 2 as X 2 + Y 2 for natural numbers Z, X, and Y.

60. N o w using one degree of freedom, which may be viewed as the external freedom

for the plane figures, the restrictions of the two-dimensional format of the plane figures

can be kept in abeyance . Here it may be relevant to note that the proof of the Pythagoras

Theorem, with the help of a right-triangle, may 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 trian

gle. 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 converse , the reconstruction of a square of the hypotenuse in terms of the squares of

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

mitted by the square as a regular body of the two-dimensional space. Further, we have

to use eight symmetr ies of the square and one degree of freedom (which the square as a

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

6 1 . As is shown above, on the two-dimensional format of OX, OY dimensional lines

we can have plane figure A B C D (which may be a square or a rectangle). This figure

wou ld have a degree of freedom of mot ion along the OZ axis. This provides us with the

possibil i ty to choose two identical figures, say A B C D and A'B'C 'D' as shown above.

The diagonal , be ing on the one-dimensional format, would have length equal to any real

number , which includes the natural numbers . The two-dimensional figure AC C'A' may

be a square on a proper choice of vert ices.

62 . It is because of the above freedom of construction for the plane figures in a three-

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

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C'A') in terms of the squares on the dimensional l ines of the format of the plane figure

A B C D and vice-versa. Though this ass ignment as such is not taken up here, i t may not

be out of context to note that the unit square would be the basic constituent which

would be used for a division process of one square into two squares or the reverse pro

cess of composi t ion of two squares into one square. The unit square (and as such any

square) has eight symmetr ies , 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

composi t ion process is possible .

6 3 . Hence we have justified the restriction of Fe rma t ' s Last Theorem to n ^ 3.

64. N o w 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

m o v i n g from two-dimensional bodies to three- and higher-dimensional bodies .

C a s e of 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

65 . Step 1: Let vo lume V = Z 3 cubic units. In other words , the cube Z is consti tuted

by Z 3 unit cubes .

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

a long 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 . Step 5: The origin point of the Cartesian frame, being a dimensionless point, when

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 vo lume V = Z 3 , but have

ended with a cube of vo lume Z(Z+1)(Z+1) which is greater than Z 3 , hence there is a

contradict ion.

Conclusion 7 3 . Therefore, the above contradict ion proves that our assumption regarding permis

sibility of the vo lume of the cube to be separated and handled l x l x l = 1 as a natural

unit is wrong . Hence X 3 cubic units cannot be treated as linear units and as such the ad

dit ion operat ion of natural numbers is not a meaningful operation. If cubic uni ts l x l x l

are treated as l inear 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 wel l , as the natural numbers addit ion operation is geometrical ly linear in nature;

the same is meaningless and i t is not applicable to the higher-dimensional units l x l ,

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

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75. Z2 ( l x l ) and Z

2 (1) are two distinct units. Similarly Z

3 ( l x l x l ) and Z

3 (1) as well

as Z4 ( l x l x l x l ) and Z

4 (1) and so on, are pairs of distinct units. The addition operat ion

which binds 1+1 is meaning less for the units l x l or l x l x l , or l x l x l x l and so on, since

( l x l ) + ( l x l ) means an operat ion in terms of which two square units are to be com

bined, whi le 1+1 = 2 is the natural numbers addition operation. Unless and until we

know how the square unit is transformable into a linear unit, the linear units addition op

eration would not be applicable. The natural numbers multiplication operation, which is

int imately connected with the addit ion operat ion, remains dependent upon the addition

operat ion only within a linear space. The moment the space stands changed from one-

dimens iona l (l inear) space to two and higher-dimensional space, the independent char

acterist ics of the mult ipl icat ion operat ion are displayed. As such, unless and until l x l is

suitably defined, the addition operat ion will remain only the operation of one-

dimens ional space.

Geometrical Continuum: Fourth-Dimensional Space

76 . Now we may take up the fourth numbe r o f the f ramed doma in s s equence

[an/2na

n-1 for n = 1, 2, 3 ... ]. For the present it may be taken by way of definition that

[a4/8a

3] is the formulat ion for the regular body of fourth-dimensional space of dimen

sional length a, content part (domain part) as a4, and the frame part as 8a

3. In the con

text we may refer to a cube of dimensional length a, content (domain part) as vo lume a3,

and frame part as surface area 6a2 as the three-dimensional regular body. Further, by

way of definition, we may take the following figure as a geometr ical presentat ion of the

fourth-dimensional regular body.

Structural Arrangement of Fourth Framed Domain

77. Algebraical ly we know that (A+B)4 = A

4 + 6A

3B + 4A

2B

2 + 6AB

3 + B

4. Geomet

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

an open interval), in terms of the above expansion, A+B = [OO'] = [OX] + (XO'] =

[OX] + [OX), when X is a rational number and is the middle point of [OO'] . If [OX] =

X, then [OO'] = 2X.

7 8 . T h e a l g e b r a i c equa l i t y [OO ' ]4 = [OX]

4 + 6 [OX ]

3 [OX) + 4 [OX ]

2 [OX)

2 +

6 [OX] [OX)3 wo rk s out the structural a r rangement of the fourth framed domain . This,

when trans lated into geometry , shall d iv ide the fourth-dimensional body into seven

teen four th-dimens ional bodies .

79 . Out of the seventeen four th-dimens ional bodies , the first, namely [OX]4, is a

comp le t e four th-dimens ional body in the sense that all the eight cubic componen t s of

the f rame 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 algebraic equality, in all we shall have five types of

d imensional 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 . Therefore, the internal structural ar rangement of the fourth-dimensional regular

body admits a break-up as f ive-dimensional b locks such that these blocks have respec

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

b locks are structurally distinct.

8 3 . N o w the general b inomial expansion of ( A + B ) N would help us conclude that we

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

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

d imensional b locks is to remain the same, so as to structure a regular body, exactly 2 N

N-dimens iona l bodies would be required.

84 . Therefore, for N 3 and Z 3 (for Z = 1 and 2 the theorem is obvious) , 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

theorem when one of X or Y is greater than the other). Let Z = X + R. Therefore, XN +

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

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

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

N o w the right side is an expression 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 . Hence , the left side cannot be equal to the the right side. With

this the geometr ical proof of the theorem stands completed.

8 5 . N o w if in the light of the above, we try to find the significance of the internal

structural a r rangement of the very first regular body l 1 , where 1 is equal to an irrational

length, then it really would be enchant ing 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 which , as a linear cont inuum, is equivalent to the field of real numbers whose sub-

field is the field of rational numbers .

86 . Obvious ly , the implicat ions are many and far reaching, particularly when they

demol i sh menta l b locks of the phys ica l wor ld and help t ranscend to four and higher-

d imens iona l spaces as a geometr ical cont inuum of spaces whose regular bodies consti

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

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