Question 627836
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Assume *[tex \LARGE \exists\ a\ \in\ \mathbb{R}\ :\ a\ \neq\ 0,\ \ a\ =\ -a]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ -a]


Add *[tex \LARGE a] to both sides:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2a\ =\ 0]


Multiply by *[tex \LARGE \frac{1}{2}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\ =\ 0]


Which contradicts the original assumption.  Therefore, reductio ad absurdum,  *[tex \LARGE a\ =\ -a\ \Leftrightarrow\ a\ =\ 0]




John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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