Is there an argument for why $\large \frac {\pi}{e}$ is rational, irrational, or trandescendal? Can the quotient of any two transcendental numbers (which are not rational multiples of each other) be rational, or at least irrational? Thanks.
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If the quotient of any two transcendental numbers, they are rational multiples of each other. – Eckhard Aug 05 '13 at 17:26
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Also, the quotient of two transcendental numbers is either going to be rational or irrational, so I don't understand what your second question is supposed to be about (I assume you're just wondering when (if at all possible) the quotient of two transcendental numbers is rational). – Andrew D Aug 05 '13 at 17:28
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@AndrewD Sorry about that, I fixed the question now. I mean to say that if the quotient is not transcendental, can it at least be irrational. – Ovi Aug 05 '13 at 17:31
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4@AndrewD The quotient of two transcendentals can be transcendental, e.g. $\frac{\sin 1}{\cos 1}$. – Lord Soth Aug 05 '13 at 17:37
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3... and the quotient of some other two transcendentals can be rational, e.g. $(2\pi)/\pi$. And the quotent of some other two transcendentals can be irrational and algebraic, e.g. $(\pi\sqrt{2})/\pi$. – GEdgar Aug 05 '13 at 17:39
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1You might want to take a look at this: http://math.stackexchange.com/questions/456097/are-pi-and-e-algebraically-independent – PersonX Aug 05 '13 at 17:43
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The quotient of two transcendentals can be anything except $0$. – André Nicolas Aug 05 '13 at 17:44
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With $e$ also $\frac{1}{e}$ is transcendental. The polynomial $(x-\pi)(x-\frac{1}{e})=x^2-(\frac{1}{e}+\pi)+\frac{\pi}{e}$ cannot have rational coefficients, hence either $\frac{1}{e}+\pi$ or $\frac{\pi}{e}$ is irrational. But indeed, the question is whether $π$ and $e$ are algebraically independent. See also https://mathoverflow.net/questions/33817/work-on-independence-of-pi-and-e.
Dietrich Burde
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