Yet another proof that 1=0

Question

Could someone explain to me where is the mistake in the following reasonning?

$$e^1=e^{\frac{2i\pi}{2i\pi}}= ((e^{i\pi})^2)^{\frac{1}{2i\pi}}= 1^{(\frac{1}{2i\pi})}=1=e^0$$ So $0=1.$

Thank you very much, sorry if it looks dumb, trying to learn by myself.

Fractional powers of negative numbers are not uniquely defined, and the "general rule" $(a^m)^n$=$a^{m×n}$ does not always work when $m$ and $n$ are not integers — J. W. Tanner, Dec 30 '19 at 18:34
Related, if not duplicate: https://math.stackexchange.com/q/397743/42969. — Martin R, Dec 30 '19 at 18:42

score 3 · Answer 1 · answered Dec 30 '19 at 18:44

3

Further to the comments, nor can you say $e^{2\pi i}=e^0\to 2\pi i=0\to 1=0$, because $e^z=e^w\not\to z=w$. So this kind of argument is doomed no matter how carefully you try to justify the $z\mapsto\sqrt[2\pi i]{z}$ step.

answered Dec 30 '19 at 18:44

J.G.

115,835

Yet another proof that 1=0

1 Answers1