Question 498474
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No importa lo que *[tex \Large a + b] es igual a: *[tex \Large (a\ -\ b)\ =\ - (b\ -\ a)] para todos los valores reales de *[tex \Large a] y *[tex \Large b], por lo tanto *[tex \Large (a\ -\ b)\ +\ (b\ -\ a)\ =\ 0] para todos los valores reales de *[tex \Large a] y *[tex \Large b].


Sorry about the Spanish, but I don't speak it.  I used Google Translate.


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