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I have algebra problem from a friend, that is 1=-1!!! because

$$-1=-1^{3}=-1^{^{\frac{6}{2}}}=\sqrt{(-1)^6}=\sqrt{1}=1$$

I can not see what is wrong with this? I will appreciate it any help.

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    $\begingroup$ Try a simpler problem $1^2=(-1)^2 \not\Rightarrow 1=-1$. $\endgroup$ – user Jan 10 '19 at 10:26
  • $\begingroup$ Notice that 1 has two roots: $1$ and $-1$. $\endgroup$ – denklo Jan 10 '19 at 10:41
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The usual properties of roots/exponents apply when the basis is (real) positive. In this case, we can't have

$$\;-1^3=(-1)^3=(-1)^{6/2}\color{red}{\stackrel {!!}=}\left[(-1)^{1/2}\right]^6$$

as $\;(-1)^{1/2}=\sqrt{-1}\;$ cannot be done within the real numbers (and this is also another reason why the above mentioned properties don't apply to complex numbers...)

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  • $\begingroup$ All $-1^3$, $(-1)^3$ and $(-1)^{6/2}$ are equal! The problem is that $a^{b/c} = \sqrt[c]{a^b}$ is not always true. $\endgroup$ – Christoph Jan 10 '19 at 10:33
  • $\begingroup$ @Christoph Agree with you, so I'll add. Thankx $\endgroup$ – DonAntonio Jan 10 '19 at 10:59
  • $\begingroup$ I'd still disagree that this is the problem. Over the complex numbers $(-1)^{1/2}$ is defined as the principal square root of $-1$, which is $i$ and $i^6=-1$. The problem is that this is different from $\left[ (-1)^6 \right]^{1/2}$, which is $1$. $\endgroup$ – Christoph Jan 10 '19 at 11:30
  • $\begingroup$ @Christoph Over the complex numbers it is false in general: $$1=\sqrt1=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=i\cdot i=-1$$ You can't do the above, certainly. We're talking here of general propieties . And in the complex number I don't know of any definition a priori of "principal or not" branches, though many times one usually choses the "usual" branches. $\endgroup$ – DonAntonio Jan 10 '19 at 12:50

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