Quote:
Originally Posted by shek Theorem: e=1 Proof:
let 2*e = f-------------->(1)
where f>0
rasing to the power 2*pi*i
(1)==>2^(2*pi*i) * e^(2*pi*i) = f^(2*pi*i)-------------->(2)
but,e^(2*pi*i) = 1 (since from complex no.s , i.e. e^(2*pi*i) =cos(2*pi)+i sin(2*pi) )
Therefore
(2)==>2^(2*pi*i) = f^(2*pi*i)
==>2=f (since power are equal bases can be equated)
Thus, from (1), we have....
e=1 
expecting contradictions..................frm u all my friends |
sry friends....
in above proof i was not able 2 show raising 2 the power..
see this proof...I have showed raising 2 the power...
here...
Quote:
2^3=8 i.e 2 power 3 = 8
3^3=27
3^2=9
pi = 22/7 = 3.14....
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