Question 199509
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And you are unclear with very good reason.  If you add *[tex \Large \frac{3}{4}] and *[tex \Large \frac{1}{3}] you get a sum greater than one, namely *[tex \Large \frac{13}{12}], so there is less than nothing left for the alleged third piece of your circle.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{3}{4} + \frac{1}{3} = \left(\frac{3}{4}\right)\left(\frac{3}{3}\right) + \left(\frac{1}{3}\right)\left(\frac{4}{4}\right) = \frac{9}{12} + \frac{4}{12} = \frac{13}{12}]


Sort of the way our Government does arithmetic, except they use bigger numbers and it is real money.



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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