In Search of a New Applicable Geometry

S. Kalimuthu

2/394 , Kanjampatti P.O , Pollachi Via , Tamil Nadu 642003 , India

Email : sona.sonasona.sona7@gmail.com , parabrammam@yahoo.in

Abstract

It is well known that great mathematical minds like Proclus , Possidonious , Playfair , Wallis , Carnot ,Gauss , Escher , Gersonides , Clavio Rome , F.K.Schweikart , F.A. Taurinus , Omar Khayyam , Saccheri , Lambert , Legendre , Bolyai , Riemann , Lobachevski, Cayley, Klein, Beltrami , Poincare and others tried their best to prove the fifth Euclidean postulate but their rigorous attempts were unsuccessful. The studies devoted to this problem gave birth to two different models of non – Euclidean geometries namely hyperbolic and elliptic. Albert Einstein applied the concepts of these geometries to formulate his general relativity theory. In the late nineteenth century mathematicians established that it is not merely difficult but impossible to deduce Euclid I from Euclid I to IV. In this work , the author re – probed the parallel postulate problem and found several new consistent findings which may give birth to a new field of science.

Key words: Classical geometry, arithmetic , algebra , unsolved physics problems , new science

MSC: 11A99 , 08C99 , 51 M04

A brief timeline of the parallel postulate

300 B. C. Euclid’s Elements were published. The fifth postulate is not self-evident, and in virtue of its relative complexity and scant intuitive appeal, in the sense that it could be proved, many mathematicians tried to present a proof of it.

300 B.C Aristotle’s linked the problem of parallels lines to the question of the sum of the angles of a triangle in his Prior Analytics.

100 B.C Diodorus, and Anthiniatus proved many different prepositions about the fifth postulate.

410-485 Proclus described an attempt to prove the parallel postulate due to Ptolemy. Ptolemy gave a false demonstration of if two parallel lines are cut by a transversal, then the interior angles on the same side add up to two right angles, which is an assertion equivalent to the fifth postulate. Proclus also gave his own (erroneous) proof, because he assumed that the distance between two parallel lines is bounded, which is also equivalent to the fifth postulate.

500’s Agh_n_s (Byzantine scholar) proved the existence of a quadrilateral with four right angles, which lead to a proof of the fifth postulate, and this was the key point of most medieval proofs of the parallel postulate, but his proof is based on an assertion equivalent to the fifth postulate.

900’s Ab_ ‘Al_ ibn S_n_ (980-1037) defined parallel straight lines as equidistant lines, which is anequivalent proposition to the fifth postulate. The Egyptian physicist Abu ‘Al_ Ibn al-Haytham(965-1041) presented a false proof for the fifth postulate, using the same idea of equidistant lines, to define parallel lines. The first attempt to prove the parallel postulate that is based on a more intuitive postulate is the theory of parallel lines of ‘Umar Khayy_m, who used the idea of to construct a quadrilateral. This same idea was latter on (18th century) used by Saccheri, and it has played a very important rule in the history of non-Euclidean geometry.

1600’s Pietro Antonio Cataldi (1548-1626), from Italy, assuming that there exist equidistant lines deduced a number of assertions from which it is already possible to prove the parallel postulate. John Wallis (1616-1703), English mathematician, also wrote a treatise dealing with the parallel postulate, where he proved the parallel postulate resting on the following postulate: to every figure, there exists a similar figure of arbitrary magnitude. Both assumption come to be equivalent to the fifth postulate.

1700’s Girolamo Saccheri (1667-1733), an Italian mathematician, made what turned out to be an important attempt to prove the parallel postulate, using a quadrilateral. In 1778 the Swiss mathematician Louis Bertrand (1731-1812) published a clever proof of the parallel postulate. But a mistake in his arguments was quickly brought to light by the Russian Emel’yanovi_ Gur’ev(1746-1813), and he, using another false conclusion, proved the parallel postulate.

1800’s In the first half of the 19th century there appeared several erroneous proofs of the parallel postulate by the Hungarian mathematician Farkas Bolyai. Friedrich Ludwig Watch r (1792-1817), a German student under Gauss, had attempted to prove the fifth postulate, and he believed that he had been successful. Bernhard Friedrich Thibaut (1775-1832) gave on other erroneous proof of the postulate. As we can notice above, during 2000 years, many mathematicians presented fake proofs, or proofs using assumptions that were equivalent to the fifth postulate, to prove it. Gauss was the first to have a clear view of a geometry independent of the fifth postulate(it was the birth of non-Euclidean geometry), and after him N.I. Lobatschewsky and Johann Bolyai. They share the honour of having made the first really systematic study of what we nowcall hyperbolic geometry. The existence of such non-Euclidean geometries proves that the fifth postulate is independent of the other ones, that is, it cannot be proved.- Ferdinand Karl Schweikark (1780-1859) developed a geometry independent of Euclid’s hypothesis (fifth postulate).

1823 Lobatschewsky thought about Imaginary geometry. Bolyai discovered a formula that is the key for all Non-Euclidean trigonometries. The most interesting of the Non-Euclidean construction given by Bolyai is that for the squaring of the circle.- The full recognition that spherical geometry is itself a kind of Non-Euclidean geometry, withoutparallels, is due to Riemann (1826-1866).

1823-1860 The acceptance of the Non-Euclidean geometry was delayed by some reasons, such as the difficult of mastering Lobatschewsky’s work written in Russian.

1860-1863 The correspondences between Gauss and Schumacher, published between 1860 and 1863, the numerous references to the works of Lobatschewsky and Bolyai, were a big step in direction to spread non-Euclidean geometries.

1871-1873 Klein suggested calling the geometries of Bolyai and Lobachewsky, Riemann, and Euclid, respectively, Hyperbolic, Elliptic, and parabolic.

1882 Henri Poincare (1854-1912) presented two models, in a half-plane and in a circle, for hyperbolic geometry.

1900’s Non-Euclidean geometries were widely spread, and the study of geometry on surfaces gave origin to differential geometry. Non-Euclidean geometry became a very useful tool for many areas, and a very wide field of research.

Case 1 Construction for case 1

Describe a quadrilateral ABCD as shown in figure 1. Join points B and D. Choose two points K

and H on BC and DC respectively. Join A and K contacting BD at R. Since points A and K lie on

the opposite sides of BD , AK can meet BD. Please note that Euclid uses this principle.

[Elements I, prop. 10 ] Similarly join A and C meeting RD at J and connect A and H touching JD

at T. Small letters denote the sum of the interior angles of triangles and quadrilaterals. Also. let

the angles sums of triangles and quadrilaterals be

ABD = a , ABT = b , ARD = c , ABJ = d , ART = e , AJD = f , ABJK= g , AJC = h , ACH = i

AHD = j , BCD = k = RJCD = l , BCHT = t , JCD = u , RJCHT = w and BCJ = a’

Euclidean Figure 1 for case 1

Results of case 1

The angles BKC , CHD , ARF , AJC , ATH , BRJ , RJT and JTD are all straight

angles and so their measures are all equal to 180 degrees. Let v be the value of

this 180 degrees (1.1)

Assuming (1), x + y + z + m = 3v + a (2.1)

2v + b = x + y + z (3.1)

y + z + m = 2v + c (4.1)

v + d = x + y (5.1)

y + z = v + e (6.1)

v + f = m + z (7.1)

x + r = v + g (8.1)

2v + h = y + q (9.1)

2v + i = p + z (10.1)

m + n = v + j (11.1)

n + p + q + r = 5v + k (12.1)

3v + l = n + p + q (13.1)

p + q + r = 3v + t (14.1)

2v + u = p + n (15.1)

p + q = v + w (16.1)

2v + a’ = q + r (17.1)

Adding (2) to (17) we get that, 2m + b + d + f + 2r + h + i + l + a’ + u =

a + c + g + j + k + t + w + e + 2v

From (8) we have 2v + 2g = 2x + 2r

Putting (3) in (2), v + a = m + b

Adding m + b = d + f

Adding 2x + 2c = 2m + 2b

(5) + 916) = (8) + (9) =( x + y + r + q ) = a’ + d = h + g

Adding (15) and (16) and using (11) we get , v + k = a’ + u

Adding the above seven relations, a + c + d + i + l + 2v = j + t + w + e + b

Adding (4) and (5) and applying (2) , y + a = c + d

v + j = m + n (10.1)

3v + t = p + q + r (13.1 )

v + w = p + q (15.1)

v + e = y + z (6.1)

2v + b = x + y + z (3.1)

p + z = 2v + i (10.1)

n + p + q = 3v + l (13.1)

Adding the above nine equations we get that , 5v + 1a = x + y + z + m + r + q

Applying (2) and (1`7) in RHS we obtain that a = a’ (18.1)

Analyzing a . a’ , (2) and (17) we get that the angle sums of triangles

ABD and CBJ are equal (19.1)

We can construct such that triangles ABC and BCD are congruent.

This implies that a = k (20.1)

Comparing (18) and (20) we have a = a’ = k (21.1)

Adding (15) and (16) and putting (12) we have , u + a’ = v + k (22.1)

Applying (21) in (22) we obtain that u = v (23.1)

Analyzing (1) and u we get that the interior angle sums of triangle JCD is 180 degrees (24.1)

Case 2:Construction for case 2

Draw triangles ABC and DBC as shown in figure 1. On AB , choose a point E. Join C and E.

Join E and D meeting BC at R. Since points E and D lie on the opposite sides of BC, Ed can

meet BC. Please note that Euclid uses this principle. [ Elements I, prop.10 ] Similarly join A

and D contacting EC at S and RC at O. Small letters denote the angle sums of triangles and

quadrilateral ORES. Also, let that a , b , c , d , e , f , g , h , I , j , k , l , t and w respectively refer

to the sum of the interior angles of triangles and quadrilaterals ACD , ACO , CDS , AED , SED

, AERO , BDE , AEC , EBC , ERC , BOSE , BCD , CDR and BOD.

A

E

B

D

CR

T

S

O

x

q

r

p

y

z

n

m

Euclidean figure 2 for case 2

Results of case 2

The angles BRO , ROC , ERD , ASO , SOD , ESC and AEB are all straight angles and so their

measures are all equal to 180 degrees. Let v be the value of this 180 degrees. (1.2)

Applying (1) , x + y + z = 2v + a (2.2)

x + y = v + b (3.2)

y + z = v + c (4.2)

m + n + p = 3v + d (5.2)

m + n = 2v + e (6.2)

p+ n = v + f (7.2)

q + r = v + g (8.2)

x + p = v + h (9.2)

y + n + q = 3v + i (10.2)

y + n = 2v + j (11.2)

n + q = v + k (12.2)

z + m + r = 2v + l (13.2)

z + m = v + t (14.2)

m + r = v + w (15.2)

Adding (2) to (15) we have , 3x + 5y + 4z +5m + 6n + 3p + 3q + 3r =

22v + a + b + c + d + e + f + g + h + i + j + k + l + t + w

Putting (4) , (6) , (8) and (9) in LHS, 2h + 2g + 3c + 4e + y + n =

2v + a + b + d + f + i + j + k + l + t + w

Applying (11) in LHS , 2h + 2g + 3c + 4e = a + b + d + f + i + k + l + t + w (16.2)

(4) + (6) = (11) + (14) = y + z + m + n = 3v + c + e = 3v + j + t, i.e c + e = j + t (17.2)

Applying (17) in (16) we obtain that , 2g + 2h + 2c + 3e + j =

a + b + d + f + i + k + l + w

2v + a = x + y + z (2.2)

v + b = x + y (3.2)

3v + d = m + n + p (5.2)

v + f = p + n (7.2)

3v + i = y + n + q (10.2)

v + k = n + q (12.2)

2v + l = z + m + r (13.2)

v + w = n + r (15.2)

y + n = 2v + j (11.2)

m + n = 2v + e (6.2)

Adding the above ten relations , 10v +2g + 2h + 2c + 3e =

2x + 2y + 2z + m + 3n + 2q + 2p +2 r

Assuming (4), (6) and (9) in RHS , 2v + 2 e = 2n , i. e n = v + e (18.2)

Applying (18) in (6) we obtain that , m = v (19.2)

Comparing (1) and m we get that the sum of the interior angles of triangle

ORD is equal to two right angles (20.2)

Case 3 Construction for case 3

Let A , B and C are the given points not lying on a straight line. Join B and C. On BC , take a

point D. Join A and D. Produce AD up to E. Join B and E ; and join C and E. On AD , choose a

point O. Join C and O. Extend CO up to G. Join G and B ; and join G and A. Join B and O.

Produce BO up to H. Join A and H ; and join C and H. Now join A and B contacting OG at F.

Since points A and B lie on the opposite sides of OG , AB can meet OG. Please note that Euclid

uses this principle. [ Elements I , prop.10 ] Similarly join A and C touching OH at J. Small letters

denote the angle sums of triangles. Also , let that a , b , c , d , e , f , g , h , i , j , k , l , t , t’ , w , a’ ,

b’ , c’ and d’ respectively refer to the sum of the interior angles of triangles and quadrilaterals

BCG , BDOG , BCF , BOG , BDOF , OBC , BEC , ABE , ABD , ABO , ABJ , ACG , ACF ,

ACO , AEC , COE , ABH and BCH.

A

F H

t q

x s r p

y o n

z m

B D C

K e1

E

Euclidean Figure 3 for case 3

Results of case 3.

The angles AFB , BDC , CJA , COF , OFG , BOJ and OJH are all straight angles and so their

measures are all equal to 180 degrees. Let v be the value of this 180 degree. (1.3)

Assuming (1) and adding, x + y + z + m = 3v + a (2.3)

v + b = x + y + z (3.3)

y + z + m = 2v + c (4.3)

v + d = x + y (5.3)

y + z = e (6.3)

v + f = m + z (7.3)

u + e’ = v + g (8.3)

3v + h = u + z + y + s (9.3)

z + y + s = 2v + i (10.3)

v + j = y + s (11.3)

y + s + r = 2v + k (12.3)

3v + l = t + s + n + r (13.3)

s + r + n = 2v + t’ (14.3)

v + w = n + r (15.3)

n + r + m + e’ = 3v + a’ (16.3)

v + b’ = m + e’ (17.3)

3v + c’ = y + s + r + q (18.3)

z + m + n + p = 3v + d’ (19.3)

Adding (2) to (19), m + b + d + z + f + e’ + h + j + w + b’ + c’ + l +( z + m + n + p) =

a + c + x + e + g + i + k + t + t’ + a’ + q + d’ + 3v

Putting (3) in (2) , v + a = m + b

Adding v + a = d + f

Applying (4) in (2) , x + c = v + a

3v + d’ = z + m + n + p

Adding the above five equations , 2v + l + z + e’ + h + j + w + b’ + c’ =

a + g + i + k + t + t’+ a’ + q

e = y + z (6.3)

v + g = u + e’ (8.3)

2v + i = z + y + s (10.3)

2v + k = y + s + r (12.3)

Applying (14) in (13) t + t’ = v + l

3v + a’ = n + r + m + e’ (16.3)

u + z + y + s = 3v + h (9.3)

y + s = v + j (11.3)

n + r = v + w (15.3)

Adding (17) and (18) , m + e’ + y + s + r + q = 4v + b’ + c’

Adding the above eleven relations we obtain that , s = v (20.3)

Comparing s and v we have that the interior angle sums of triangle

ASF is 180 degrees (21.3)

Case 4

A

y z D

J

m C

x o

B r n

q

H p

F

D1

E

A1

Euclidean figure 4 for case 4

1. Construction [ For case 4 ]

Draw a triangle ABC. Choose three points J , O and D on AB , BC and CA respectively. Join O and A , O and J and O and D. Produce AO , JO and DO up to A’ , E and D’ respectively. Join E and C. Join B and E contacting DD’ at H and AA’ at F. Since points B and E lie on the opposite sides of AA’ and DD’ , BE can meet AA’ and DD’ . Please note that Euclid uses this principle. [ Elements I , prop. 10 ] Small letters denote the angle sums of triangles and quadrilaterals. Also ,let a , b , c , d , e , f , g , h , u , j , k , l , t , u , w , and a’ respectively refer to the sum of the interior angles of triangles and quadrilaterals ABC , ABOD , AJOC , ABO , ACO , BCE , OCEH , BOE , OCEF , BOF , ABF , AFEC , OECD , OFEC , BJOH and AJOD.

2. Results

The angles AJB , BOC , CDA , BHF , HFE , AOF , JOE and HOD are all straight angles and

so their measures are all equal to 180 degrees. Let v be the value of this 180 degree (1.4)

Assuming (1) and adding , x + y + z + m = 3v + a (2.4)

x + y + z = v + b (3.4)

y + z + m = v + c (4.4)

x + y = v + d (5.4)

m + z = v + e (6.4)

n +p + q + r = 3v + f (7.4)

n + p + q = v + g (8.4)

p + q + r = 2v + h (9.4)

p + n = i (10.4)

r + q = v + j (11.4)

3v + k = x + y + r + q (12.4)

2v + l = z + m + n + p (13.4)

t = m + n (14.4)

u = p + n (15.4)

w = x + r (16.4)

a’ = y + z (17.4)

Adding (2) to (16) , x + 2y + 2z + m + 2p + 3q + r + k + l + t + u + w + a’ =

9v + a + b + c + d + e + f + g + h + i + j

3v + a = x + y + z + m (2.4)

Adding (3) and (4) and pitting (2), b + c = v + a + y + z (2a.4 )

2v + h = p + q + r (9.4)

Adding the above four equations , p + 2q + k + l + t + u + w + a’ =

5v + a + d + f + g + i + j

Adding (9) and (10) , v +i + j = p + n + r + q

v + g = n + p + q (8.4)

v + d = x + y (5)

Adding (5) and (6) and applying (2) d + e = v + a

x + y + r + q = 3v + k (12.4)

z + m + n + p = 2v + l (13.4)

3v + a = x +y + z + m (2.4)

Adding the above eight relations , t + u + w + a’ = 5v + n + a + f (18.4)

Adding (14),(15),(16) and (17), x + y + z + m + p + 2n + r = t + u + w + a’ (19.4)

Adding (18) and (19) we get that , x + y + z + m + p + n + r = 5v + a + f

Putting (2) in LHS , p + n + r = 2v + f (20.4)

Assuming (200 in (7) we obtain that q = v (21.4)

Analyzing (1) and q we have that the interior angle sums of

triangle HOF is 180 degree (22.4)

Case 5

Construction for case 5Draw a triangle ABC as shown in figure 1.Take two points E and O on AB and BC respectively. Join A and O and produce AO up to D. Join E and O and extend EO up to F. Join E and D contacting BO at R. Since points E and D lie on the opposite sides of BO , ED can meet BO. Please note that Euclid uses this principle.[ Elements I , prop. 10 ] Similarly join C and D touching OF at H. Also , join B and D. Small letters denote the angle sums of triangles. Also let that a , b , c , d , e , f , g , h , i , k , l , t , u and w respectively refer to the sum of the interior angles of triangles and quadrilaterals ABC , ABO , AERC , BEO , SERO , AEOC , BCD , RCD , BDHO , COD , ROHD , BOD , BED , AED and ACD.

A

E z

m

B x y o C

R

q n

p

r H

D

F

Euclidean Figure 5 for case 5

Results for case 5The angles AEB , BRO , ROC , ERD , EOH , AOD and CHD are all straight

angles and so their measures are all equal to degrees. Let us assume that v be the

value of this 180 degree (1.5)

Assuming and adding x + y + z + m = 3v + a (2.5)

2v + b = x + y + z (3.5)

y + z + m = v + c (4.5)

v + d = x + y (5.5)

y + z = e (6.5)

f = m + z (7.5)

n + p + q + r = 3v + g (8.5)

2v + h = n + p + q (9.5)

p + q + r = v + i (10.5)

v + j = n + p (11.5)

p + q = k (12.5)

v + l = r + q (13.5)

v + t = x + r (14.5)

2v + u = z + y + q (15.5)

2v + w = m + n + p (16.5)

Adding equations (2) to (16) we have , b + f + h + j + l + t + u + w + 4v =

a + c + e + g + i +2 n + k + x + q (17.5)

Adding (14) , (15) and (16)

and then putting (2) and (8) we have v + a + g = t + u + w (18.5)

Adding (3) , (7) ,( 9) , (11) and (13) we have ,

2n + 2p + 2q + r + x + y + 2z + m = 6v + b + f + h + j + l (19.5)

(4) +(6) +(10) + (12 ) = 2v + c + e + i + k = 2y + 2z + m + 2p + 2q + r (20.5)

Adding the above four equations we obtain that v = y (21.5)

Analyzing v and y we obtain that the angle sums of triangle ORE is

equal to a straight angle (22.5)

Case 6 Construction for case 6 [ see the below figure ]

A

X y z m

B D E F C

Euclidean figure 6 for case 6

Let ABC be the given triangle. Take points S , E , and F on BC. Join A and D ; join A and E

and join A and F. Small letters x , y , z and m respectively refer to the angle sums of triangles

ABD , ADE , AEF and AFC. Also , let a , b , c , d and e respectively refer to the sum of the

interior angles of triangles ABE , ADF , AEC , ABF and ADC.

Results of case 6

The angles BDE , DEF and EFC are all straight angles and so their measures are all equal

To 180 degrees. Let v be the value of this 180 degree. (1.6)

Assuming (1) and adding , x + y = v + a (2.6)

y + z = v + b (3.6)

z + m = v + c (4.6)

a + z = v + d (5.6)

y + c = v + e (6.6)

Let us transform the elements of equations (2) to (6) in to the following four matrices:

A = { 0 0 0 0 -v z } (7.6)

B = { x -y m -z v -z } (8.6)

C = { a -b c -d 0 v } (9.6)

D = { e 0 -v v v -a } (10.6)

A + B + C + D = { x+a+e -y-b m+c-v v-z-d v v -a } (11.6)

A – B = { -x y -m z -2v 2z } (12.6)

D – C = { e-a b -v-c -v –d v -a-v } (13.6)

Adding these three, 2A + 2D = { 2e 0 -2v -2d 0 2z-2a } (14.6)

From (7.6) and (10.,6)2A +2D = { 2e 0 -2v 2v 0 2z -2a } (15.6)

Equating the above two we get that 2v = -2d (16.6)

It is well known that in geometry, minus theta is the vertically opposite angels.

Since vertically opposite angles are equal we get from (17) that d = v (17.6)

Comparing (1.6) and (17.6) we obtain that the interior angle sums of triangle

ABF is 180 degrees. (18.6)

Case 7

Construction for case 7 [ same construction for case 6 ]

[ To avoid confusion , commas are given ]

A x-y , v-e

z-d , m-v (1.7)

B = a-e , d-b

0 , -v (2.7)

C = v-x , e-v

d-v , v-m (3.7)

D = 0 , e-v

d-v , x-m (4.7)

(7) + (10) gives , A + D = x – y , 0

z-v , x-v (5.7)

(7) + (8) gives , C + A = v-y , 0

z-v , 0 (6.7)

(10) – (7) gives , D – A = y-x , 2e-2v

v+z-2d, 2m-v-x (7.7)

(10) – (8) gives , D – B = e-a , b+e-v-d

d-v , x+m-v (8.7)

(9) – (9) gives , B-C = x+a-v-e , v+d-b-e

v-d , m-2v (9.7)

Adding the above five relations, A + 3D = x-y , 2e – 2v

3z-2d-v , x+2m-3v (10.7)

3D = 0 , 2e-3v

3d-3v , 3x-3m (11.7)

(7) + (17) gives , A + 3D = x-y , 2e-2v

2d+z-3v , 3x-2m-v (12.7)

Equating the elements of (16) and (18) we obtain , 2d+z-3v = 3z-2d-v

i.e 4d = 2z + 2v = 2d = v + z (13.7)

and 3x – 2m – v = x + 2m + -3v = 2x + 2v = 4m = x + v = 2m (14.7)

Putting (1) in (4) we have , a + z = v + d (4a)

2d = v + z (15.7)

(4a) + (19) gives , a + d = 2v (16.7)

x+ v = 2m (17.7)

Let us assume that in triangle ABC ,

AB = AC , AD = AF and AE is perpendicular to AC .

This implies that x = m (18.7)

Comparing (20) and (21) we get that x = v (19.7)

Analyzing x and v we obtain that the interior angle sum of triangle

ABD is 180 degree (20.7)

Case 8

Construction for case 8

Draw a triangle ABC. Choose three points J , O and D on AB , BC and CA respectively. Join O and A , O and J and O and D. Produce AO , JO and DO up to A’ , E and D’ respectively. Join E and C. Join B and E contacting DD’ at H and AA’ at F. Since points B and E lie on the opposite sides of AA’ and DD’ , BE can meet AA’ and DD’ . Please note that Euclid uses this principle. [ Elements I , prop. 10 ] Small letters denote the angle sums of triangles and quadrilaterals. Also ,let a , b , c , d , e , f , g , h , u , j , k , l , t , u , w , and a’ respectively refer to the sum of the interior angles of triangles and quadrilaterals ABC , ABOD , AJOC , ABO , ACO , BCE , OCEH , BOE , OCEF , BOF , ABF , AFEC , OECD , OFEC , BJOH and AJOD.

2. Results for case 8

The angles AJB , BOC , CDA , BHF , HFE , AOF , JOE and HOD are all straight angles and

so their measures are all equal to 180 degrees. Let v be the value of this 180 degree (1.8)

Assuming (1) and adding , x + y + z + m = 3v + a (2.8)

x + y + z = v + b (3.8)

y + z + m = v + c (4.8)

x + y = v + d (5.8)

m + z = v + e (6.8)

n +p + q + r = 3v + f (7.8)

n + p + q = v + g (8.8)

p + q + r = 2v + h (9.8)

p + n = i (10.8)

r + q = v + j (11.8)

3v + k = x + y + r + q (12.8)

2v + l = z + m + n + p (13.8)

t = m + n (14.8)

u = p + n (15.8)

w = x + r (16.8)

a’ = y + z (17.8)

Adding (2) to (16) , x + 2y + 2z + m + 2p + 3q + r + k + l + t + u + w + a’ =

9v + a + b + c + d + e + f + g + h + i + j

3v + a = x + y + z + m (2.8)

Adding (3) and (4) and pitting (2), b + c = v + a + y + z (2a )

2v + h = p + q + r (9.8)

Adding the above four equations , p + 2q + k + l + t + u + w + a’ =

5v + a + d + f + g + i + j

Adding (9) and (10) , v +i + j = p + n + r + q

v + g = n + p + q (8.8)

v + d = x + y (5)

Adding (5) and (6) and applying (2) d + e = v + a

x + y + r + q = 3v + k (12.8)

z + m + n + p = 2v + l (13.8)

3v + a = x +y + z + m (2.8)

Adding the above eight relations , t + u + w + a’ = 5v + n + a + f (18.8)

Adding (14),(15),(16) and (17), x + y + z + m + p + 2n + r = t + u + w + a’ (19.8)

Adding (18) and (19) we get that , x + y + z + m + p + n + r = 5v + a + f

Putting (2) in LHS , p + n + r = 2v + f (20.8)

Assuming (200 in (7) we obtain that q = v (21.8)

Analyzing (1) and q we have that the interior angle sums of

triangle HOF is 180 degree (22.8)

A

y z D

J

m C

x o

B r n

q

H p

F

D1

E

A1

Euclidean figure 8 for case 8

Case 9

Construction for case 9

Draw a triangle ABC as shown in figure 1.Take two points E and O on AB and BC respectively. Join A and O and produce AO up to D. Join E and O and extend EO up to F. Join E and D contacting BO at R. Since points E and D lie on the opposite sides of BO , ED can meet BO. Please note that Euclid uses this principle.[ Elements I , prop. 10 ] Similarly join C and D touching OF at H. Also , join B and D. Small letters denote the angle sums of triangles. Also let that a , b , c , d , e , f , g , h , i , k , l , t , u and w respectively refer to the sum of the interior

angles of triangles and quadrilaterals ABC , ABO , AERC , BEO , SERO , AEOC , BCD , RCD , BDHO , COD , ROHD , BOD , BED , AED and ACD.

2. Results for case 9The angles AEB , BRO , ROC , ERD , EOH , AOD and CHD are all straight

angles and so their measures are all equal to degrees. Let us assume that v be the

value of this 180 degree (1.9)

Assuming and adding x + y + z + m = 3v + a (2.9)

2v + b = x + y + z (3.9)

y + z + m = v + c (4.9)

A

E z

m

B x y o C

R

q n

p

r H

Euclidean figure 9 for case 9

D

F

v + d = x + y (5.9)

y + z = e (6.9)

f = m + z (7.9)

n + p + q + r = 3v + g (8.9)

2v + h = n + p + q (9.9)

p + q + r = v + i (10.9)

v + j = n + p (11.9)

p + q = k (12.9)

v + l = r + q (13.9)

v + t = x + r (14.9)

2v + u = z + y + q (15.9)

2v + w = m + n + p (16.9)

Adding equations (2) to (16) we have , b + f + h + j + l + t + u + w + 4v =

a + c + e + g + i +2 n + k + x + q (17.9)

Adding (14) , (15) and (16)

and then putting (2) and (8) we have v + a + g = t + u + w (18.9)

Adding (3) , (7) ,( 9) , (11) and (13) we have ,

2n + 2p + 2q + r + x + y + 2z + m = 6v + b + f + h + j + l (19.9)

(4) +(6) +(10) + (12 ) = 2v + c + e + i + k = 2y + 2z + m + 2p + 2q + r (20.9)

Adding the above four equations we obtain that v = y (21.9)

Analyzing v and y we obtain that the angle sums of triangle ORE is

equal to a straight angle (22.9)

Case 10

Construction for case 10

Let A, B and C are the three given points. Join A and B; join B and C; and join C and A. On BC

take two points D and E. Join A and D. Join A and E. Let X, Y and Z denote the sum of the

interior angles of triangles ABD, ADE and AEC respectively. Also let that A’, B’, and C’

respectively refer to the sum of the interior angles in triangles ABE, ADC and ABC.

A

X Y Z

B D E CFigure 1 (Euclidean)

A’ = sum of the interior angles of triangle ABE

B’ = sum of the interior angles of triangle ADC

C’ = sum of the interior angles of triangle ABC

Results for case 10

The angles BDE ands DEC are straight angles and so their measures are equal

to180 degrees. Let V be the value of this 180 degree (1.19)

Using (1) , X + Y = V + A ‘ (2.10)

i.e. X = { V + A’ – Y } (2a.10)

X + B’ = A’ + Z = V + C ‘ (3.10)

i.e. A’ = { X + B’ – Z } (3a.10) Also, A’ = { V + C’ – Z } (3b)

X + Y + Z = 2V + C ‘ (4.10) i.e. X = { 2V + C’ – Y – Z } (4a.10)

Let us assume that eqns. (2a),(3a),(3b) and (4a) denote sets.

Intersection of the sets A and B, denoted A ∩ B, is the set of all objects that

are members of both A and B. The intersection of {1, 2, 3} and {2, 3, 4} is the set {2, 3} .

Let us assume that equations (2a) , (3b) and (4a) are sets.

Considering equations (2a) and (3b) and applying the intersection law of set theory we obtain that X ∩ A’ = { V } (5.10)

Taking equations (3b) and (4a) and assuming interjection law of set theory we have that, X ∩ A’ = { - Z } (6.10) Comparing (5) and (6) we get that V = - Z (7.10)

It is well known that in geometry minus theta refers to the vertically opposite angles.

Since vertically opposite angles are equal, (7) implies that V = Z (8.10) i.e. The sum of the interior angles of triangle AEC is a straight angle (9.10)

Case 11

Construction and proof for case 11

NA’, NB ’and NC’ are the segments of a sphere S whose north pole is N. Choose a point A on

NA’. With center N, radius NA, describe an arc cutting AB’ at B and AC’ at C.

So, NA = NB = NC (1.11)

Take a point F on NA. With center N, radius NF draw an arc meeting NB at E

and NC at D. So, NF = NE = ND (2.11)

From (1) in triangle NAC, angle NAB = angle NCB ( 3.11)

And in triangle, NBC, angle NBC = angle NCB (4.11)

From (1.3) and (1.4) we obtain that angles NAB=NBC=NCB = 90 degree (5.11)

Similarly from (2) we can show that angles, NFE= NEF= NED=NDE = 90 o (6.11)

From (1.5) and (1.6) we get that the sum of the interior angles of spherical

quadrilaterals BCDE, ABEF and ACDF is 3600 (7.11)

Spherical figure 11 for case 11

N

F DE

A

B

C

A’ B’ C’

D

A

R T

B C E F

S

Figure 12 for case 12(Spherical) S Figure 12a for case 12 (Spherical)

Case 12

Construction for case 12

Let ABC and DEF are spherical equilateral triangles such that all the three interior

angles are right angles and DE = 2 AB. Let R , S , and T are the mid points of DE ,

EF and FD respectively Join B and T , T and S and S and R.

Results for case 12

By side angle side correspondence triangles ABC , DRT , ERS and STF are congruent.

So all the interior angles of these equilateral triangles are right angles

Using ( 1 ) in figure 7a, we get that the sum of the angles,

DRT + ERS = ESR + FST =DTR + STF = 180 degree (1.12) But DRT + TRS + ERS = ESR + RST +FST = DTR + RTS + STF = 180o (2.12)

Comparing (5.1) and (5.2) we get t a contradiction that the angles

RST=STR=TRS=0 and the angle sum of triangle RST is zero. (3.12)

Also , since triangles ABC and DRT are congruent the corresponding angles

are also equal. Since each angle of triangle ABC is a right angle , we consequently

get that the angles ERT and FTR are equal and their value is 90 degree. From this

we obtain that the sum of the interior angles of spherical quadrilateral ERTF is equal

to 360 degrees. (4.12) Discussion

The results of the above 12 cases are given below:

The interior angle sums of triangle JCD is 180 degrees (24.1)

The sum of the interior angles of triangle ORD is equal to two right angles (20.2)

The interior angle sums of triangle ASF is 180 degrees (21.3)

The interior angle sums of triangle HOF is 180 degree (22.4)

The angle sums of triangle ORE is equal to a straight angle (22.5)

The interior angle sums of triangle ABF is 180 degrees. (18.6)

The interior angle sum of triangle ABD is 180 degree (20.7)

The interior angle sums of triangle HOF is 180 degree (22.8)

The angle sums of triangle ORE is equal to a straight angle (22.9)

The sum of the interior angles of triangle AEC is a straight angle (9.10)

The sum of the interior angles of spherical quadrilaterals

BCDE, ABEF and ACDF is 3600 (7.11)

The interior angles of spherical quadrilateral ERTF is equal

to 360 degrees. (4.12)

Since we have deduced the above twelve results without assuming the fifth Euclidean postulate , they establish the parallel postulate. [ 1 – 5 ] But the mere existence of non Euclidean geometries and their physical applications demonstrate that the parallel postulate is independent. More over Beltrami , Cayley, Klein , Poincare and others showed that it is impossible to deduce Euclid V from Euclid I to IV. But the above twelve findings are consistent. Cases 11 and 12 are the author’s masterpieces. So , these inventions convey a mathematical message that there is something hidden treasure .Only further studies can unlock the mystery. This may give rise to a new field of mathematics.

Due to the rapid expansion of the frontiers of physics and engineering, the demand for higher-level mathematics is increasing yearly. Theoretical Physics is the formulation and mathematical development of fundamental physical theories, such as mechanics, relativity, cosmology and thermodynamics. The Physics is a vast subject and includes areas such as: Astrophysics includes study of the origins and evolution of the solar system, making use of data collected from international telescopes and satellite observatories. Mathematics and physics are fundamental to many aspects of modern life. This includes technology, such as the computer, the laser, the compact disc, nanotechnology, space travel, mobile phones, MRI (magnetic resonance imaging)… the list is endless! Government, industry and commerce also utilize the detailed analytical and modelling capability of mathematics and statistics to underpin many activities. But despite these real facts, there are many unsolved and challenging problems in science particularly in physics. To know the ultimate nature of this universe is a burning cosmological problem. In order to solve the current unsolved problems, the origin of a new mathematical field is required.

Mathematical concepts are widely applied in mechanics of particles and systems , mechanics of deformable solids , fluid mechanics , optics, electromagnetic theory , classical thermodynamics, heat transfer , quantum theory , statistical mechanics, structure of matter , relativity and gravitational theory , astronomy and astrophysics , geophysics , bio physics , systems theory, control theory , biology and other natural sciences , game theory, economics, social and behavioral sciences , land survey , architecture and all the engineering filed. Without the applications of mathematics no science and even no arts can develop and flourish. There are many great unsolved and challenging problems in physics such as quantum gravity, understanding the nucleus, fusion energy, climate change, turbulence, glassy materials, high-temperature superconductivity, solar magnetism, complexity, gravitational waves and their detection , neutron stars and pulsars. supernova stars , black holes. cosmic strings , quasars and galactic nuclei. formation of galaxies , the problem of dark matter hidden mass and itsdetection , the origin of super high-energy cosmic rays , gamma-ray bursts. hyper novae , neutrino physics and astronomy , neutrino oscillations , non conservation of CP invariance and consciousness. In order to solve and to know the ultimate reality of Nature , the creation of new mathematical fields is a must.

Physics is an area of science where many branches of mathematics have been directly applied. Nature seems to obey 'mathematical rules' rather than acting whimsically. In other words, it seems that natural laws can be expressed in terms of mathematics To a physicist, mathematics is a toolbox. Before attacking a particular problem, you should have the necessary tools for the job. There are some tools that should be in any physicist's toolbox, but as they specialize, they will add extra tools needed for the specific problems at hand. Each new development in physics often requires a new branch of mathematics.. The language of physics is mathematics but physics is applied mathematics. The following connections have been widely agreed: Classical mechanics – Calculus and geometry , Electromagnetism - Vector calculus , General relativity - Spherical geometry, differential geometry and differential equations , Quantum field theory - Matrices, group theory , differential equations and probability, Superstring theory - Knot theory. There are many challenging and unsolved physical problems. In order to get solutions for these burning problems , a new mathematical idea should be created

“If you have not enjoyed Euclid in your youth , you are probably not made for a scientific career.” Einstein

6. Acknowledgments

The author wishes to thank the late professor of mathematics Dr. Palaniappan Kaliappan, Mathematics Department, Nallamuthu Gounder Mahalingam College, Pollachi, Tamil Nadu 642001, India for his kind encouragement for the preparation of this paper.

References

[1] Kalimuthu , S. , (2010) , Pretty Algebra , I. Journal of Appld. Maths. &

Applications, 2 (1) , 29 - 33

[2] http://www.ripublication.com/aa/aav2n1_6.pdf

[3 ] http://www.scipub.org/fulltext/jms2/jms242122-123.pdf

[4] http:// doi:10.1016/j.camwa 2009.07.048

[5] http://dx.doi.org/10.1016/j.aml.2010.08.006