Question 199356
<font face="Garamond" size="+2">


BUZZZ! I'm sorry, but your answer is incorrect.  But thank you for playing.  We have some lovely parting gifts on your way out.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ a\cdot 1 = a\ \forall a\ \in\ \R]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{12}{12} = 1]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{\ \frac{2}{3}\ }{\frac{3}{4}} = \left(\frac{\ \frac{2}{3}\ }{\frac{3}{4}}\right)\left(\frac{12}{12}\right) = \frac{8}{9}]


But *[tex \Large \frac{8}{9}] is rational because 8 and 9 are both integers.  Since a number cannot be both rational and irrational at the same time, and *[tex \Large \frac{8}{9}] and *[tex \Large \frac{\ \frac{2}{3}\ }{\frac{3}{4}}] are demonstrably the same number, *[tex \Large \frac{\ \frac{2}{3}\ }{\frac{3}{4}}] must be rational.


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