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01-12-2017, 02:19 PM

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Originally Posted by WhiteWalker View Post
මේ ඔප්පු කිරීම වැරදියි බං

ඔතන රවුම් කල්ල තියෙන තැන බලන්න
when n=2 , [2] comes from [1]
so they are connected and see a,b,c are common for the right angle triangle while n vary
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01-12-2017, 03:34 PM

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Originally Posted by luxmen View Post
when n=2 , [2] comes from [1]
so they are connected and see a,b,c are common for the right angle triangle while n vary
you are still saying while n=2 , we can get eqn [2] from eqn [1] . so ultimately eqn [2] valid where n=2 but not for any other values.
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01-12-2017, 03:56 PM

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Reason ??
Quote:
Originally Posted by luxmen View Post
when n=2 , [2] comes from [1]
so they are connected and see a,b,c are common for the right angle triangle while n vary
ඔතන කල්ල තියෙන්නෙ සමීකරණ දෙක්ක අරං දෙකේම LHS < RHS නිසා RHS1 = RHS2 කියල අරං එහෙම ගන්න බෑ
5<10
5< 7 වුනා කියල 7 = 10 වෙන්නෑ
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01-12-2017, 04:20 PM

owanam , pransha wage dan...
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01-12-2017, 11:32 PM

Quote:
Originally Posted by WhiteWalker View Post
ඔතන කල්ල තියෙන්නෙ සමීකරණ දෙක්ක අරං දෙකේම LHS < RHS නිසා RHS1 = RHS2 කියල අරං එහෙම ගන්න බෑ
5<10
5< 7 වුනා කියල 7 = 10 වෙන්නෑ
5<7 and 5<10 ---there is no connection between your two inequalities.For example---x<10 and x^2 <100 ------there is a connection between two inequalities ,
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01-12-2017, 11:35 PM

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Originally Posted by Sataninhell View Post
you are still saying while n=2 , we can get eqn [2] from eqn [1] . so ultimately eqn [2] valid where n=2 but not for any other values.
Yes , n=2 proved in PDF1-----------THEN PLEASE SEE PDF2 FOR n>2
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for n>2 - 01-12-2017, 11:38 PM

http://vixra.org/pdf/1601.0114v1.pdf
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01-12-2017, 11:38 PM

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01-13-2017, 01:21 AM

Originally Posted by WhiteWalker
ඔතන කල්ල තියෙන්නෙ සමීකරණ දෙක්ක අරං දෙකේම LHS < RHS නිසා RHS1 = RHS2 කියල අරං එහෙම ගන්න බෑ
5<10
5< 7 වුනා කියල 7 = 10 වෙන්නෑ


Quote:
Originally Posted by luxmen View Post
5<7 and 5<10 ---there is no connection between your two inequalities.For example---x<10 and x^2 <100 ------there is a connection between two inequalities ,
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 ???

Last edited by mag123; 01-13-2017 at 01:22 AM.
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01-13-2017, 06:24 AM

Quote:
Originally Posted by mag123 View Post
Originally Posted by WhiteWalker
ඔතන කල්ල තියෙන්නෙ සමීකරණ දෙක්ක අරං දෙකේම LHS < RHS නිසා RHS1 = RHS2 කියල අරං එහෙම ගන්න බෑ
5<10
5< 7 වුනා කියල 7 = 10 වෙන්නෑ




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 THE RIGHT ANGLE TRIANGLE. YOU CAN NOT FIX VALUES FOR,a,b,c,n AS YOU WANT.-------FOR EXAMPLE---THIS VALUES MATCH FOR A RIGHT ANGLE TRIANGLE---a=3,b=4,c=5,n=2-------now check
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