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Multiplication is iterated addition; exponentiation is iterated multiplication, tetration is iterated exponentiation, call all of those as different grades of iterated arithmetical operators.

Is there a known $n$-iterated arithmetical operator such that $\mathbb C$ is not closed under it?

Zuhair
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    Already raising an arbitrary complex number to an arbitrary complex power is not well-defined. – hardmath Apr 30 '19 at 22:52
  • Indeed, raising a nonreal complex number to a power is well defined only when the exponent is in $\Bbb Z$. – Lubin May 01 '19 at 03:33
  • @Lubin, but why then it is said that the complex numbers are closed under exponentiation? – Zuhair May 01 '19 at 08:39
  • http://mathworld.wolfram.com/ComplexExponentiation.html –  May 01 '19 at 14:58
  • They’re closed all right, it’s just that the operation can not be unambiguously defined. – Lubin May 01 '19 at 16:51
  • @Lubin, what do you mean by "unambiguously define" can you please clarify this point? – Zuhair May 01 '19 at 17:32
  • Well, there’s no way to define $i^x$ for $x\in\Bbb R$ in such a way that you always have $i^\alpha\cdot i^\beta=i^{\alpha+\beta}$. – Lubin May 01 '19 at 19:32
  • @Lubin but there is a rule that would reduce to that when $i,\alpha, \beta$ are all in $\mathbb R^+$, so there is no real problem. – Zuhair May 02 '19 at 05:07
  • But by $i$ I meant $\sqrt{-1}$. The reason it all works as you say above is that there is a log defined for positive reals making $\log(ab)=\log a +\log b$ always. But there is no such thing for the nonzero complex numbers. – Lubin May 02 '19 at 16:13
  • I think it's time to point out that you can get the same answer by iteration, not that they are exactly equivalent. –  May 03 '19 at 14:08
  • @RoddyMacPhee, what do you exactly mean? can you clarify? – Zuhair May 03 '19 at 16:28
  • addition is defined in terms of the sucessor function, and multiplication can be but really doesn't need to be. –  May 03 '19 at 16:42

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