Question 584659
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You can't.  In fact, you can easily prove that, in general, *[tex \LARGE \frac{a}{b}\ \neq\ \frac{b}{a}]


Let *[tex \LARGE a\ =\ 1] and let *[tex \LARGE b\ =\ 2], then *[tex \LARGE \frac{1}{2}\ \neq\ \frac{2}{1}]


And, in fact, *[tex \LARGE \frac{a}{b}\ =\ \frac{b}{a}] if and only if *[tex \LARGE a\ =\ b] because 1 is the only real number equal to its own reciprocal.


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