- #1
tom92373
- 3
- 0
I was playing around with euler's formula the other day and found this odd proof which says 2ipi=0. I know this is obviously wrong, but what was the step that went wrong? The "proof" goes like this:
start with e^(i*pi)+1=0
so -e^(i*pi)=1
multiply both sides by e e(-e^(i*pi))=e
factor a (-1) (-1)(e)(e^(i*pi))=e
add exponents (-1)(e^(i*pi+1))=e
take ln of both sides ln((-1)(e^(i*pi+1)))=1
use properties of logs ln(-1) + ln(e^(i*pi+1))=1
take ln of -1 i*pi + ln(e^(i*pi+1))=1
properties of logs i*pi + (i*pi+1)ln(e)=1
take ln(e) i*pi + i*pi+1=1
subtract 1 i*pi + i*pi=0
combine like terms 2*pi*i=0 ...
this even works in the original formula:
e^(i*2*pi)=cos(2*pi)+isin(2*pi)
e^(i*2*pi)=1
e^(i*2*pi)=e^0
i*2*pi=0 ...
start with e^(i*pi)+1=0
so -e^(i*pi)=1
multiply both sides by e e(-e^(i*pi))=e
factor a (-1) (-1)(e)(e^(i*pi))=e
add exponents (-1)(e^(i*pi+1))=e
take ln of both sides ln((-1)(e^(i*pi+1)))=1
use properties of logs ln(-1) + ln(e^(i*pi+1))=1
take ln of -1 i*pi + ln(e^(i*pi+1))=1
properties of logs i*pi + (i*pi+1)ln(e)=1
take ln(e) i*pi + i*pi+1=1
subtract 1 i*pi + i*pi=0
combine like terms 2*pi*i=0 ...
this even works in the original formula:
e^(i*2*pi)=cos(2*pi)+isin(2*pi)
e^(i*2*pi)=1
e^(i*2*pi)=e^0
i*2*pi=0 ...