4 ms·
Yeah, this proof seems to work. I was thinking of the case where you attack $e^2$ directly, instead of going after both $e$ and $e^{-1}$.
by siegelzero 10y ago
Yeah, this proof seems to work. I was thinking of the case where you attack $e^2$ directly, instead of going after both $e$ and $e^{-1}$.
- pmalynin 10y agoIt can be shown by clever application of Taylor Expansion that e^2 is irrational.