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Since we know that the value of $e$, $i$, and $\pi$ are irrational reals, how about $$e^{i+\pi}\;?$$ Is it still irrational (that is, not a Gaussian rational)? The problem make me curious until now.

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    $\begingroup$ Obviously, since it is not even real. $\endgroup$ Jul 5, 2013 at 7:06
  • $\begingroup$ Please tell us what you think $e^{i+\pi}$ is $\endgroup$
    – jimjim
    Jul 5, 2013 at 7:10
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    $\begingroup$ Well, here en.wikipedia.org/wiki/Irrational_number , here thefreedictionary.com/irrational+number and here mathworld.wolfram.com/IrrationalNumber.html , just to mention a few, says otherwise, @Arjang . In all this "irrational" number is considered a real number. I'd also go with this definition, btw. $\endgroup$
    – DonAntonio
    Jul 5, 2013 at 8:36
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    $\begingroup$ I misread @Arjang 's comment: too many negatives too close to each other for me. Ithought he said "irrational does not depend on being real" ... $\endgroup$
    – DonAntonio
    Jul 5, 2013 at 8:38
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    $\begingroup$ $i$ is not irrational. It's not even real. $\endgroup$ Jul 5, 2013 at 12:23

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Building off of Clive's answer, if $e^\pi \cos 1$ and $e^\pi \sin 1$ were both rational numbers, then so would be their quotient $\tan 1$. But $\tan 1$ is irrational as a consequence of the Lindemann–Weierstrass theorem, because that theorem implies that $e^{2i}$ is transcendental, and because $e^{2i}=\dfrac{i-\tan 1}{\tan 1+i}$. Therefore $e^{\pi+i}$ cannot have both its real and imaginary parts rational.

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If you mean rational in the usual sense $-$ as a subset of the reals $-$ then $e^{i+\pi}$ is certainly not rational since it is not real. But the question of whether it's a Gaussian rational $-$ that is, its real and complex parts are rational $-$ may not be easy to prove. Notice that $$e^{i+\pi} = e^{\pi}\cos 1 + i e^{\pi}\sin 1$$

Given irrational numbers $\alpha$ and $\beta$ it's often quite hard to determine whether $\alpha^{\beta}$ and $\alpha \beta$ (and $\alpha + \beta$) are rational or irrational. For example, it is not currently known whether $e\pi$ is rational or irrational.

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    $\begingroup$ It would suffice to show that $\tan 1$ is irrational. I guess that is true but don't know how to prove it and don't have a reference. (Incidentally, it is known that $e^\pi$ is irrational.) $\endgroup$ Jul 5, 2013 at 8:56
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$e^i=\cos1+i\sin1$. (By Euler's Formula)

So $e^{i+\pi}=e^\pi\cos1+ie^\pi\sin1$, which is non-real and, of course, irrational.

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    $\begingroup$ en.wikipedia.org/wiki/… irrationality does not depend on not being real. $\endgroup$
    – jimjim
    Jul 5, 2013 at 8:23
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    $\begingroup$ @Arjang: Apparently you are using a different definition of rational than ᴊ ᴀ s ᴏ ɴ. That does not mean that JASON's interpretation is incorrect. It is common for rational to refer only to elements of the set of rational numbers, $\mathbb Q$, whereas you seem to prefer that it means an element of $\mathbb Q+i\mathbb Q$. $\endgroup$ Jul 5, 2013 at 8:35
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    $\begingroup$ @Arjang: I disagree. That wikipedia article talks about rational points on a curve, that is points such that all their coordinates are rational. The set of rational numbers is most certainly a subset of the reals. $\endgroup$ Jul 5, 2013 at 8:36
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    $\begingroup$ @Arjang: There is a subtle issue of terminology here. A rational number is necessarily real. A rational point in the complex plane is one with rational coefficients. $\endgroup$ Jul 5, 2013 at 8:36
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    $\begingroup$ Sorry about being a bit blunt in my previous comment. But a Gaussian rational is simply not necessarily rational. In the same sense elements of $\mathbb{Q}(\sqrt{-3})$ would be called Eisensteinian rationals. The algebraic integers of these two fields are widely known as Gaussian (resp. Eisensteinian) integers. All basic theory of algebraic number fields. But we never call all elements of algebraic number fields "rationals". $\endgroup$ Jul 5, 2013 at 12:47

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