My Simple Attempted Proof On Fermat's Last Theorem
- From: jiahao_anti-addictgamer@xxxxxxxxxxx
- Date: 14 Mar 2007 23:18:54 -0700
Dearest fellow members ,
I have made the har decision to show my proof to everyone , who can
comment on it and lure out flaws.
I agree I may not have breach defination of what called a mathematical
argument .I am in all ears to listen to your suggestions . The proof
follows as it deems fit :
Proof On Fermat's Last Theorem
Part i)
In this part , we are going to proof on the basis that Pythagorean
Triples will never work for FLT .
The proof , is as follows .
Pythagoras Theorem states that: Fermat's Last Theorem
states that:
a^2 + b^2 = c^2-------( 1 ) x^n + y^n = z^n:no
solutions ------ ( 2 )
-----------------------------------------------------------------------
First we look at c^2 , then at z ^n.
For c^2 to be in the nth power ,
RHS = c^2 (c^(n-2)) = c^n
As follows ,
LHS = a^2 + b^2 ( c^(n-2) ) =a^2(c^(n-2) ) + b^2( c^(n-2) )
We now apply proof by contradiction to the proof.
Asumming that FLT is true ,
Then,
a^2(c^(n-2) ) + b^2( c^(n-2) ) =a^n + b^n
But,
for a^2(c^(n-2) ) + b^2( c^(n-2) ) =a^n + b^n,
c^(n-2) must be equal to a^(n-2) and b^(n-2)
which is clearly not the case , i.e., there is an contradiction
therefore if
a^2(c^(n-2) ) + b^2( c^(n-2) ) =/= a^n + b^n ,
Pythagorean Triples aka pt(x) cannot fit into the equation , i.e. ,
the proof is complete .
Part ii)
The Idea Of A "Forced" Pythagorean Triple
By rearranging the consistent p-triple of a class, we conclude the
rearrange we an DNA that varies according to the relevant result.
These newly revolutionized triples , also called the function f of f
class p-triple i.e. , pt(f) , these are made up of imaginary numbers
and rational numbers that are varied to the value of c .
---------------------------------------------------------
Our aim is now to prove that pt(f) will not work for FLT , which will
indirectly prove that FLT is true .
Say we test the rational numbers , m/v and r/s .
Apply proof by contradicting again ,
If FLT is false , then FLT must have a solution through the
multiplication of pt(x) or pt(f).
x^n + y^n = c^n (m/v)^2 + (r/s)^2 = c^2
------- ( 3 )
As again , we multiply the RHS and LHS ( From eqn ( 3 )
RHS = c ^2(c^(n-2) = c^n
For the LHS , we apply a special technique that will be the utmost
foundation of the proof .
LHS = (m/v)^2((x^n + y^n)/ ((m/v)^2 + (r/s)^2 ) + (r/s)^2((x^n + y^n)/
((m/v)^2 + (r/s)^2 ) =
= m^2(x^n + y^n ) r^2(x^n + y^n)
----------------------- +
----------------------
m^2 + r^2v^2/s^2 s^2m^2 + r^2
Assuming FLT is false , then,
m^2(x^n + y^n ) r^2(x^n + y^n)
----------------------- + ---------------------- =x
^n + y ^n
m^2 + r^2v^2/s^2 s^2m^2 + r^2
then,
m^2 r^2
----------------------- + ---------------------- =1
m^2 + r^2v^2/s^2 s^2m^2 + r^2
m^2 + (mr)^2 = m^4s^2 +(mr)^2 + (rvm)^2
m^2 = m^4s^2 + (rvm)^2
But in logic , the above equation cannot exist .
Hence , m^2 =/= m^4s^2 + (rvm)^2
Conclusion:There is no solution for FLT by Case 1
FLT holds true for Case 1 where pt(f) holds a consistency of rational
numbers.
Case 2: pt(f) with consistency of imaginary numbers cannot work for
Fermat's equation .
Say two imaginary numbers i ,sqrt(a) , sqrt(b).
sqrt(a)^2 + sqrt(b)^2 = sqrt(c)^2 -------- ( 4 )
Multiplying the LHS and RHS ( From eqn ( 4 )
RHS= sqrt(c)^2(sqrt(c)^(n-2) = sqrt(c)^n
Also applying the technique again ,
LHS = sqrt(a)^2(sqrt(c)^2(sqrt(c)^(n-2) / sqrt(a)^2 + sqrt(b)^2 )
+
sqrt(b)^2(sqrt(c)^2(sqrt(c)^(n-2) / sqrt(a)^2 +
sqrt(b)^2 )
Again , applying proof by contradiction ,
Assuming FLT is false ,
Then ,
the equation :
sqrt(a)^2(sqrt(c)^2(sqrt(c)^(n-2) / sqrt(a)^2 + sqrt(b)^2 )
+
sqrt(b)^2(sqrt(c)^2(sqrt(c)^(n-2) / sqrt(a)^2 + sqrt(b)^2 ) = x^n +
y^n ,
must exist.
Following the proof by contradiction,
sqrt(a)^2(sqrt(c)^2(sqrt(c)^(n-2) / sqrt(a)^2 + sqrt(b)^2 )
+
sqrt(b)^2(sqrt(c)^2(sqrt(c)^(n-2) / sqrt(a)^2 + sqrt(b)^2 ) =
( sqrt(a^2c^n) / sqrt(a^4) + sqrt(ab)^2 ) + ( sqrt(b^2c^n / sqrt
(ab)^2 + sqrt (b^4) ) = (ac^(n/2) / a^2 + (ab)) +( bc^(n/2) / (ab) +
b^2)
=(c^(n/2) / a + b) + ( c^(n/2) / a + b)
=2(c^(n/2) / a + b)
= c^n
2(c^(n/2) = ac^n + bc^n = ( a + b)(c^(n/2) (c^(n/2)
Comparing coefficents ,
For the above equation to exist i.e. , for Case 2 to work on Fermat's
equation ,
Then ,
( a + b)(c^(n/2) = (a+b)( sqrt(c^n)
Which again , in logic , the above equation cannot exist because
a,b,c,x,y, are all whole numbers .
Hence ,
( a + b)(c^(n/2) =/= (a+b)( sqrt(c^n)
So,
sqrt(a)^2(sqrt(c)^2(sqrt(c)^(n-2) / sqrt(a)^2 + sqrt(b)^2 )
+
sqrt(b)^2(sqrt(c)^2(sqrt(c)^(n-2) / sqrt(a)^2 + sqrt(b)^2 ) = x^n +
y^n,
the above equation will be forbidden to exist .
Hence , we can conclude that pt(f) does not hold true for Fermat's
equation.
Further more , from the proof before , we can come to a final
conclusion that
pt(f) and pt(x) do not hold for Fermat's equation i.e. , FLT is true .
--------------------------------------------------------------------------------------------
Proof prepared and created by Sim Jia Hao .
Date of completion of proof :14 / 3 / 2007
Copyright@Sim Jia Hao .
CONSTRUCTIVE COMMENTS PLEASE .
Thank you ,
John Howard Sim
15 / 3 / 2007
.
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