Question 465348
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There are no two integers such that *[tex \Large \pi] is equal to the quotient of those two integers.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \not\exist\ p\ \in\ \mathbb{Z}\text{ and }\ \not\exist\ q\ \in\ \mathbb{Z}\ |\ \pi\ =\ \frac{p}{q}]


Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pi\ \in\ \mathbb{J}\ =\ \mathbb{R}\,\setminus\,\mathbb{Q}]


In other words, *[tex \Large \pi] is irrational.


Furthermore:


There is no polynomial equation with rational coefficients such that *[tex \Large \pi] is a root.  Therefore *[tex \Large \pi] is trancendental, which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pi\ \in\ \mathbb{R}\,\setminus\,\mathbb{A}]


Where *[tex \Large \mathbb{A}] is the set of all real numbers that are roots of polynomial equations with rational coefficients.


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