ෆර්මාගේ අවසන් ගැටළුව

[luxmen]

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 ----------- (3)

kohomada kiyane inequality form ekaka LHS samana una kiyala RHS samana wennama oni kiyala ??

methanin ehata me sadanaya karana eke therumak naha. oya widiyata 3 equation eka liyanna bahane .

oba n=2 awasthawata pamanak meya sadanaya karanawa nam , nawatha n sadaha wena agayan yodanna baha, mokada e awasthawa sathya wanne n=2 awasthawata pamanak wana nisa.

n is variable while a,b,c constant nedha?when n=2 then c<a+b , -----now see, equation [1] c^n/2<a^n/2 +b^n/2,SEE THERE SAME ---c--a---b[to when n=2]. Although n is different ,c,a,b same to n=2, so if n=2 ,n=3,n=4,n=5,n=6---------but c,a,b[constant relative to n] are same to n=2. SO FOR ANY VALUE OF n , c<a+b ,so c^n/2<[a+b]^n/2 for any value 0f n

 

[luxmen]

c,a,b are same, although n is different

 

[Sataninhell]

c,a,b are same, although n is different

i was asking about following eqn .

So-----a^n/2+b^n/2=(a+b)^n/2
(a+b)^n/2 =a^n/2+b^n/2 ----------- (3)

 

[luxmen]

i was asking about following eqn .

So-----a^n/2+b^n/2=(a+b)^n/2
(a+b)^n/2 =a^n/2+b^n/2 ----------- (3)

------I give you same answer, n=2 ta c<a+b nisa n= any value also c<a+b because same c,a,b for any value of n

 

[luxmen]

so c^n/2 <[a+b]^n/2 for any value of n

 

[luxmen]

it comes from inequality [ 1]

 

[luxmen]

so left side is equal right side should be equal . Because same inequality. [2] comes from [1]

 

[mag123]

so left side is equal right side should be equal . Because same inequality. [2] comes from [1]

Ok , what you basically says is this as below.

Originally Posted by luxmen
----------------MAMA MEA LIYALA THIYENA DEYA GANA AVADANAYA YOMU KARANNA.----------INEQUALITY [2] COMES FROM INEQUALITY [1].SO LEFT SIDE IS EQUAL , RIGHT SIDE SHOULD BE EQUAL.

So if it is true then

2 < 1+2 ....(1)
2 < 3+2 ....(2)

So according to your deduction ; SO LEFT SIDE IS EQUAL , RIGHT SIDE SHOULD BE EQUAL.

1+2 = 3+2 (3) ???

 

[luxmen]

Ok , what you basically says is this as below.

Originally Posted by luxmen
----------------MAMA MEA LIYALA THIYENA DEYA GANA AVADANAYA YOMU KARANNA.----------INEQUALITY [2] COMES FROM INEQUALITY [1].SO LEFT SIDE IS EQUAL , RIGHT SIDE SHOULD BE EQUAL.

So if it is true then

2 < 1+2 ....(1)
2 < 3+2 ....(2)

So according to your deduction ; SO LEFT SIDE IS EQUAL , RIGHT SIDE SHOULD BE EQUAL.

1+2 = 3+2 (3) ???

I have to repeat same answer. YOUR (2) DOES NOT COME FROM YOUR (1].ANOTHER WAY TO SAY, THERE IS NO CONNECTION BETWEEN YOUR (1) AND (2).---------BUT SEE MY PROOF ,THERE IS A CONNECTION.

 

[mag123]

Ok , what you basically says is this as below.

Originally Posted by luxmen
----------------MAMA MEA LIYALA THIYENA DEYA GANA AVADANAYA YOMU KARANNA.----------INEQUALITY [2] COMES FROM INEQUALITY [1].SO LEFT SIDE IS EQUAL , RIGHT SIDE SHOULD BE EQUAL.

So if it is true then

2 < 1+2 ....(1)
2 < 3+2 ....(2)

So according to your deduction ; SO LEFT SIDE IS EQUAL , RIGHT SIDE SHOULD BE EQUAL.

1+2 = 3+2 (3) ???

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

Ok from your equations ; that is (2) comes from (1),
Just put c=2 , a=3 and b=4 , n=4

then,
(1) -> 4 < 9+16
(2) -> 4< (3+4)^2 =49

So according to you SO LEFT SIDE IS EQUAL , RIGHT SIDE SHOULD BE EQUAL.

so 9+16 = 49 ???

 

[Sataninhell]

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

Ok from your equations ; that is (2) comes from (1),
Just put c=2 , a=3 and b=4 , n=4

then,
(1) -> 4 < 9+16
(2) -> 4< (3+4)^2 =49

So according to you SO LEFT SIDE IS EQUAL , RIGHT SIDE SHOULD BE EQUAL.

so 9+16 = 49 ???

It seems like a contradiction , he is trying to prove contradiction and hence the theorem.

above case is valid for n=2 only.

then,
(1) -> 4 < 3+4 = 7
(2) -> 4< (3+4)^1 =7

Ultimately following formation is wrong,
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

it should be rearranged like,
From (1)---- c^n/2<a^n/2+b^n/2
From(2) ----c^n/2<(a+b)^n/2

For n=2 ---- a^n/2+b^n/2=(a+b)^n/2

---

somewhere in top, you can find following statement , which is again not true ,
In c^n/2<a^n/2+b^n/2
c,a,b are common and n/2 is variable.

n is not a variable for this contradiction . n is already assumed as integer 2 .

If the presentation formatted according to mathematical notations , then it would have more attention . Other than that the method has no flaw.
Good luck .

 

[mag123]

It seems like a contradiction , he is trying to prove contradiction and hence the theorem.

above case is valid for n=2 only.

then,
(1) -> 4 < 3+4 = 7
(2) -> 4< (3+4)^1 =7

Ultimately following formation is wrong,
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

it should be rearranged like,
From (1)---- c^n/2<a^n/2+b^n/2
From(2) ----c^n/2<(a+b)^n/2

For n=2 ---- a^n/2+b^n/2=(a+b)^n/2

---

somewhere in top, you can find following statement , which is again not true ,
In c^n/2<a^n/2+b^n/2
c,a,b are common and n/2 is variable.

n is not a variable for this contradiction . n is already assumed as integer 2 .

If the presentation formatted according to mathematical notations , then it would have more attention . Other than that the method has no flaw.
Good luck .

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

Yes it's only valid when n=2

 

[luxmen]

I have to repeat same answers again and again. a,b,c are constants relatively to n. So although n is 2 or 3 or 4 or ------ a,b,c are same. FOCUS ABOUT RIGHT ANGLE TRIANGLE GIVEN. LENGTH OF ITS FOOT. IT IS NOT C . BUT IT IS c^n/2 commonly. and a^n/2 and b^n/2 also. See there a,b,c are common[constants] while n is variable.

 

[luxmen]

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

Ok from your equations ; that is (2) comes from (1),
Just put c=2 , a=3 and b=4 , n=4

then,
(1) -> 4 < 9+16
(2) -> 4< (3+4)^2 =49

So according to you SO LEFT SIDE IS EQUAL , RIGHT SIDE SHOULD BE EQUAL.

so 9+16 = 49 ???

YOU HAVE FORGOTTEN THAT IT SHOULD BE RIGHT ANGLE TRIANGLE.SO YOU CAN NOT FIX VALUES AS YOU WANT.

 

[priyankaH]

YOU HAVE FORGOTTEN THAT IT SHOULD BE RIGHT ANGLE TRIANGLE.SO YOU CAN NOT FIX VALUES AS YOU WANT.

I think what Fermat Theorem says is except n=2 , you will not find any value of n to hold equation a^n+b^n=c^n is true.

So as long as n>2 , you will not find values a,b, and c to hold that equation true.

So you can't assume square triangle or any thing your answer only valid when n=2

 

[luxmen]

I think what Fermat Theorem says is except n=2 , you will not find any value of n to hold equation a^n+b^n=c^n is true.

So as long as n>2 , you will not find values a,b, and c to hold that equation true.

So you can't assume square triangle or any thing your answer only valid when n=2

Please see my proof with PDF from the beginning.

 

[luxmen]

http://vixra.org/pdf/1601.0053v1.pdf

 

[luxmen]

I started with n . It is a common value. I did not start with n =2 specially. My second PDF ,you can understand that when n tend to greater than 2 when a + b tend to its minimum value [c].

 

[luxmen]

http://vixra.org/pdf/1601.0114v1.pdf

 

[coolshano]

niyamai machan