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I would like to ask a fundamental question with regards to the imaginary number and it is something many beginners are told are wrong, but I would like to seek a rigorous proof of why it is wrong. It is a question I faced when my student asked me this. Take for example, $(-1)^{1/6}$. We can compute this in 2 different ways: $$(-1)^{1/6}=[(-1)^{2}]^{1/12}=1$$ or $$(-1)^{1/6}=[(-1)^{1/3}]^{1/2}=(-1)^{1/2}=i$$ I understand both the methods above are wrong, and the typical response is that you cannot square $-1$. Is there a rigorous proof as to why this method is flawed? Perhaps using abstract algebra or Galoise Theory?

Thank you!

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marked as duplicate by Simply Beautiful Art, Moishe Kohan, Dietrich Burde, egreg, Trevor Gunn Jun 16 '17 at 20:12

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    $\begingroup$ Your result is a ''rigorous proof'' that $(-1)^6$ cannot be calculated this way. $\endgroup$ – Emilio Novati Jun 16 '17 at 13:30
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    $\begingroup$ For a complex number $z$ and a non-integer rational $r$, $z^r$ is typically not defined. That is because given $n\in \Bbb{N}^*$, and $z\in Bbb{C}^*$ there are precisely $n$ complex numbers $w$ such that $w^n = z$. Therefore there isn't a unique number that you could call $z^{\frac{1}{n}}$ $\endgroup$ – Max Jun 16 '17 at 13:30
  • $\begingroup$ No need for Galois theory here ! Just write explicitely what $x^{ab}$ means when $x\in\mathbb{C}$, and see whether it means the same as $(x^a)^b$... $\endgroup$ – Evargalo Jun 16 '17 at 13:30
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    $\begingroup$ One remark to be made here. You can always square -1. This is not the problem. As others mentioned, the problem is that $(x^a)^b \neq x^{ab}$ for complex numbers. The other problem is of course that $(-1)^{1/2}$ has multiple solutions. $\endgroup$ – mlk Jun 16 '17 at 13:36
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Under the common conventions, $z=(-1)^{1/6}$ is the solution of $z^6+1=0$ closest to the positive real half axis.

In the same way, $z=((-1)^2)^{1/12}$ is the solution of $z^{12}-1=0$ closest to the positive half-axis.

As $z^{12}-1=(z^6-1)(z^6+1)$, the second solution set contains the first, however as a larger set the selection of "the" root can (and does) differ.

It is this selection of one among several equally valid roots that breaks all the nice arithmetic laws that hold for roots of positive numbers.

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