This entry was posted on Saturday, December 29th, 2007 at 11:33 pm and is filed under Comics. You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.

e^(i*pi)=-1 so e^(2*i*pi)=1 and therefore one of the solutions to the equation x^(2*i*pi)= 1 is x=e. Another solution is x=1. So you contruct the equation using one solution (e) and then present another solution (1). The flaw is to assume that these two solutions must be identical.

(Explicitly, let z=e^iv. Then, z^(2ipi)=e^(-2piv). For v=0 you get z=1 and for v=-1/(2pi) you get z=e.)

In fact, if w=e^n for any whole-numbered n, then w solves the equation x^(2*i*pi)= 1.

The last step is flawed. 1^ 1/(2i*pi) is not exclusively 1 (i.e. there are other roots).

third step is false

e^(i*pi)=-1 so e^(2*i*pi)=1 and therefore one of the solutions to the equation x^(2*i*pi)= 1 is x=e. Another solution is x=1. So you contruct the equation using one solution (e) and then present another solution (1). The flaw is to assume that these two solutions must be identical.

(Explicitly, let z=e^iv. Then, z^(2ipi)=e^(-2piv). For v=0 you get z=1 and for v=-1/(2pi) you get z=e.)

In fact, if w=e^n for any whole-numbered n, then w solves the equation x^(2*i*pi)= 1.