ෆර්මාගේ අවසන් ගැටළුව

luxmen · Jan 8, 2016

Persius said:
methana krla thiyenne n=2 thiyena gunayak n>2 ne kiyala pennana ekane.

mulinma n=2 kiyala aragena ai n/2 variable kiyanne??? wichalanaya wenne nene. n=2 . n/2 =1

n=2 wenkota terms 3 kin liyana puluwan. n>2 ehema be . ow eka hari.den aththtama ekada methana ? uththare?
a+b ^n ekak aran ai? me proof eka kiyawddi prashnei mekai kisi galapeemak ne wage denenawa. mage danum mattame awulak wennathi eka

Start ekedi n=2 kiyala mama gatthe naha. Sadarana equation ekak gatthe c^n = a^n +b^n meke n nodanna term ekaki. Eka nisa sadarana equation ekaki. n valata sadarana value ekak try kala.

luxmen · Jan 8, 2016

Namuth mea proof eke expose nokarapu kotasak mama laga thiyenava. Eya danne , Fermat saha Mama Pamani.

luxmen · Jan 9, 2016

07sanjeewakaru said:
maru...
habai aulak thibboth thiyenne methana........

From (1)---- c^n/2<a^n/2+b^n/2
From(2) ----c^n/2<(a+b)^n/2
So-----a^n/2+b^n/2=(a+b)^n/2
(a+b)^n/2 =a^n/2+b^n/2
(a+b)^n=(a^n/2+b^n/2)^2

------Trigonometry mathak kara ganna. a^n/2 and b^n/2 and c n/2 kiyanne right angletriangle ekaka padha. c^n/2 vikarrnaya venava. MEHI THIYENA ASAMANATHA[<] VALIN ADAHAS VENNE NORMAL ASAMANATHAVAYAK [<] NEVE. C ^n/2 < a ^n/2 + b ^n/2 kiyanne VIKARRNAYE [C^n/2] MAXIMUM LIMIT EKA a^n/2 + b^n/2 kiyana ekai. FOR EXAMPLE a <10 saha a MAXIMUM VALUE eka 10 vita, a <10 athara venasak thiyenava.For example a=2 nam a<10---- a < 9 -----a<8------a< 7 -----a <6------a< 5 -----a< 4----a< 3---viya haka. NAMUTH a < 10 kiyanne 10 , a ,vala maximum value eka nam 2< 10 pamanai select kala hakke.

luxmen · Jan 9, 2016

mag123 said:
you wrote in your proof

From (1)---- c^n/2<a^n/2+b^n/2 ---(a)
From(2) ----c^n/2<(a+b)^n/2 ---(b)
So----- a^n/2+b^n/2=(a+b)^n/2 ---(c) ???

(a) and (b) are inequalities so how can you say a^n/2+b^n/2=(a+b)^n/2 just because left sides are equal?

It's just like saying
2< 2+4
2< 5+7

so 2+4 = 5+ 7 ????

------Trigonometry mathak kara ganna. a^n/2 and b^n/2 and c n/2 kiyanne right angletriangle ekaka padha. c^n/2 vikarrnaya venava. MEHI THIYENA ASAMANATHA[<] VALIN ADAHAS VENNE NORMAL ASAMANATHAVAYAK [<] NEVE. C ^n/2 < a ^n/2 + b ^n/2 kiyanne VIKARRNAYE [C^n/2] MAXIMUM LIMIT EKA a^n/2 + b^n/2 kiyana ekai. FOR EXAMPLE a <10 saha a MAXIMUM VALUE eka 10 vita, a <10 athara venasak thiyenava.For example a=2 nam a<10---- a < 9 -----a<8------a< 7 -----a <6------a< 5 -----a< 4----a< 3---viya haka. NAMUTH a < 10 kiyanne 10 , a ,vala MAXIMUM VALUE eka nam, 2< 10 pamanai select kala hakke.

luxmen · Jan 9, 2016

Avastha dekedima MAXIMUM values enne. Triangle eka diha baluwama vatahenava. Avastha dekedima MAXIMUM values salaka balenne. C^n/2 MAXIMUM VALUE ekai thiyenne. Dekak thibiya nohaka. So ----a^n/2 +b^n/2 = [a+b]^n/2

07sanjeewakaru · Jan 9, 2016

luxmen said:
Avastha dekedima MAXIMUM values enne. Triangle eka diha baluwama vatahenava. Avastha dekedima MAXIMUM values salaka balenne. C^n/2 MAXIMUM VALUE ekai thiyenne. Dekak thibiya nohaka. So ----a^n/2 +b^n/2 = [a+b]^n/2

kohomada awastha dekenma ekama limit eka denawa kiyanne..... :confused:

ethana thama wadagathma tharkaya.......oya samanakireema matha thama proof ekama rada pawathinne.........
mokada asamanatha dekakin samanathawak ganne....
ethenta hoda pahadili kireemak oona wei.........

eth meka maru........
ohoma thaniyama karanan marai......
oya othana newei inta oone sirawata.... :yes:

බෙන් 10 · Jan 9, 2016

bump
rep++

luxmen · Jan 9, 2016

07sanjeewakaru said:
kohomada awastha dekenma ekama limit eka denawa kiyanne.....
ethana thama wadagathma tharkaya.......oya samanakireema matha thama proof ekama rada pawathinne.........
mokada asamanatha dekakin samanathawak ganne....
ethenta hoda pahadili kireemak oona wei.........

eth meka maru........
ohoma thaniyama karanan marai......
oya othana newei inta oone sirawata....

Thanks, and Oba educated kenek kiyala penava. MEMA ASAMANATHAVAYAN[INEQUALITIES] DEKAMA RIGHT ANGLE TRIANGLE EKATA CONNECTED. EKA NISA HODIN PARIKKSHA KARA BALANNA. MEMA AVASTHA DEKEDIMA UNIQUE OPPORTUNITY[ONE OPPORTUNITY] THIYENNE. *****NATHNAM RIGHT ANGLE EKA KADENAVA.*****EKA NISAI RIGHT SIDE EKA SAMANA VENNE.

luxmen · Jan 9, 2016

ithuru tika Mama expose karanna inne, open Meeting ekakadi. E kivve, for n> 2 , there are no positive integers for a,b,c . MEMA ITHURU EASY PROOF TIKA*****FERMA DANNAVA*****MAMA DANNAVA*****. FERMA TA KARANA GARU KIRIMAK VASAYEN, MAMA ITHIRI TIKA IDIRIPATH KARANNAM, OPEN PUBLIC MEETING EKAKADI.

mag123 · Jan 9, 2016

Originally Posted by mag123

you wrote in your proof

From (1)---- c^n/2<a^n/2+b^n/2 ---(a)
From(2) ----c^n/2<(a+b)^n/2 ---(b)
So----- a^n/2+b^n/2=(a+b)^n/2 ---(c) ???

(a) and (b) are inequalities so how can you say a^n/2+b^n/2=(a+b)^n/2 just because left sides are equal?

It's just like saying
2< 2+4
2< 5+7

so 2+4 = 5+ 7 ????
------Trigonometry mathak kara ganna. a^n/2 and b^n/2 and c n/2 kiyanne right angletriangle ekaka padha. c^n/2 vikarrnaya venava. MEHI THIYENA ASAMANATHA[<] VALIN ADAHAS VENNE NORMAL ASAMANATHAVAYAK [<] NEVE. C ^n/2 < a ^n/2 + b ^n/2 kiyanne VIKARRNAYE [C^n/2] MAXIMUM LIMIT EKA a^n/2 + b^n/2 kiyana ekai. FOR EXAMPLE a <10 saha a MAXIMUM VALUE eka 10 vita, a <10 athara venasak thiyenava.For example a=2 nam a<10---- a < 9 -----a<8------a< 7 -----a <6------a< 5 -----a< 4----a< 3---viya haka. NAMUTH a < 10 kiyanne 10 , a ,vala MAXIMUM VALUE eka nam, 2< 10 pamanai select kala hakke.

Last edited by luxmen; Today at 03:58 AM.

=====================================================

From (1)---- c^n/2<a^n/2+b^n/2 --- (a)
From(2) ----c^n/2<(a+b)^n/2 --- (b)
So----- a^n/2+b^n/2=(a+b)^n/2 ---(c) ???

Still you haven't explain how a^n/2+b^n/2=(a+b)^n/2 from equations (a) & (b) which is based on your further proof.

luxmen · Jan 9, 2016

C^n/2 saha C kiyanne rightangle triangles vala Vikarrnayange lengths. a^n/2,b^n/2 and a,b kiyanne ithiri padha deke lengths. inequalities ave meva asuren. Inequalities dekema C^n/2 thiyenava.

mag123 · Jan 9, 2016

luxmen said:
C^n/2 saha C kiyanne rightangle triangles vala Vikarrnayange lengths. a^n/2,b^n/2 and a,b kiyanne ithiri padha deke lengths. inequalities ave meva asuren. Inequalities dekema C^n/2 thiyenava.

Let’s say for simplicity D=c^n/2 , A= a^n/2 , B=b^n/2 , T=(a+b)^n/2
Then
D < A+B ---(a)
D < T ---(b)
Then you say A+B= T from above two??

luxmen · Jan 9, 2016

mag123 said:
Let’s say for simplicity D=c^n/2 , A= a^n/2 , B=b^n/2 , T=(a+b)^n/2
Then
D < A+B ---(a)
D < T ---(b)
Then you say A+B= T from above two??

YES BECAUSE WE HAVE COME ACROSS *RIGHT ANGLE TRIANGLE* . DO NOT FORGET RIGHT ANGLE TRIANGLE.INEQUALITIES TAKEN FROM RIGHT ANGLE TRIANGLE.

luxmen · Jan 9, 2016

WE HAVE COME ACROSS RIGHT ANGLE TRIANGLE. SO C^n/2 < a^n/2+b^n/2 has totally unique values. It is unique[one] . Otherwise right angle breaks.[then it is not a right angle tri angle]

luxmen · Jan 9, 2016

C^n/2----C-----a^n/2-----b^n/2------a----b--------all are foots of right angle triangle

DasunDe · Jan 9, 2016

gooooooooood

luxmen · Jan 10, 2016

mag123 said:
Originally Posted by mag123
you wrote in your proof

From (1)---- c^n/2<a^n/2+b^n/2 ---(a)
From(2) ----c^n/2<(a+b)^n/2 ---(b)
So----- a^n/2+b^n/2=(a+b)^n/2 ---(c) ???

(a) and (b) are inequalities so how can you say a^n/2+b^n/2=(a+b)^n/2 just because left sides are equal?

It's just like saying
2< 2+4
2< 5+7

so 2+4 = 5+ 7 ????
------Trigonometry mathak kara ganna. a^n/2 and b^n/2 and c n/2 kiyanne right angletriangle ekaka padha. c^n/2 vikarrnaya venava. MEHI THIYENA ASAMANATHA[<] VALIN ADAHAS VENNE NORMAL ASAMANATHAVAYAK [<] NEVE. C ^n/2 < a ^n/2 + b ^n/2 kiyanne VIKARRNAYE [C^n/2] MAXIMUM LIMIT EKA a^n/2 + b^n/2 kiyana ekai. FOR EXAMPLE a <10 saha a MAXIMUM VALUE eka 10 vita, a <10 athara venasak thiyenava.For example a=2 nam a<10---- a < 9 -----a<8------a< 7 -----a <6------a< 5 -----a< 4----a< 3---viya haka. NAMUTH a < 10 kiyanne 10 , a ,vala MAXIMUM VALUE eka nam, 2< 10 pamanai select kala hakke.

Last edited by luxmen; Today at 03:58 AM.

=====================================================

From (1)---- c^n/2<a^n/2+b^n/2 --- (a)
From(2) ----c^n/2<(a+b)^n/2 --- (b)
So----- a^n/2+b^n/2=(a+b)^n/2 ---(c) ???

Still you haven't explain how a^n/2+b^n/2=(a+b)^n/2 from equations (a) & (b) which is based on your further proof.

--------MATHEMATICALLY EXPLANATION-------[2] COMES FROM [1]. ------ SO comes from [a]. -----now right side should be equal mathematically. Because left side is equal[both inequalities come from one inequality]

luxmen · Jan 10, 2016

INEQUALITY [2] COMES FROM INEQUALITY [1]. SO IF LEFT SIDE IS EQUAL , THEN RIGHT SIDE SHOULD BE EQUAL .

luxmen · Jan 10, 2016

http://vixra.org/abs/1601.0053

KPZ · Jan 10, 2016

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