# Rigorous proof of the argument that one cannot square $-1$ in complex number problems [duplicate]

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!

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

• Your result is a ''rigorous proof'' that $(-1)^6$ cannot be calculated this way. – Emilio Novati Jun 16 '17 at 13:30
• 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 $z^{\frac{1}{n}}$ – Max Jun 16 '17 at 13:30
• 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 $(x^a)^b$... – Evargalo Jun 16 '17 at 13:30
• 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. – mlk Jun 16 '17 at 13:36

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.