e=1......see and believe ur eyes

shek · #1 12-01-2007, 05:07 AM

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

shek · #2 12-02-2007, 01:36 PM

friends want contradictions frm u.....

believe-me · #3 12-20-2007, 05:01 PM

Hi..I think
combining real & complex no.s is wrong...

shek · #4 12-20-2007, 05:06 PM

y not buddy ??
we can use it....there is no theory that we cannot use both at a time ...

shek · #5 12-20-2007, 05:12 PM

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....