Where is my mistake ? (In the field of real numbers)
$$ (-1)^3=-1 \to \sqrt[3]{-1}=-1 $$
$$\sqrt[3]{-1}=-1=(-1)^{\frac{1}{3}}=(-1)^{\frac{2}{6}}=\sqrt[6]{(-1)^2}=\sqrt[6]{(1)^2}=1$$
$$-1=^?1 $$
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1Easier $-1=(-1)^{2/2}=\sqrt{(-1)^2}=1$. But also $(-1)^{2/2}=(\sqrt{-1})^2$... ooops. – A.Γ. Nov 26 '16 at 10:55
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9This question has been asked here approximately $200$ times. – barak manos Nov 26 '16 at 10:59
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2For the record, the law $a^{bc}=(a^b)^c$ is only valid when $a$ is real and positive and $b$ is real. See this answer You implicitly used this law without satisfying the condition $a>0$ as $(-1)^{\frac26}=((-1)^2)^{\frac16}$, which is wrong. – Marc van Leeuwen Nov 26 '16 at 12:07
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That is the same error as in $1=\sqrt{1}=\sqrt{(-1)^2}=-1$
That is based on the fact: $1^\frac{1}{2} = \sqrt{1} = \pm 1$
The same is true for $1^\frac{1}{6} = \sqrt[6]{1} = \pm 1$
So you get $-1=\pm 1$ and that is true.
Etoplay
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The mistake lies in assuming that $(1)^{\frac{1}{6}}=1$, this is because, $1$ has two sixth roots in the field of real numbers, $1,-1$
vidyarthi
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No, $1$ being real and positive, $1^{\frac16}=1$ is perfectly well defined and unique. Nobody ever construes $e^{1/3}$ to be a (possibly!) non-real number or $e^{1/2}$ to be possibly negative. – Marc van Leeuwen Nov 26 '16 at 12:13
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@MarcvanLeeuwen But, arent there 2 sixth roots of unity in the field of reals? – vidyarthi Nov 27 '16 at 17:53
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Yes, but the set of $n$-th roots of unity is not produced by raising $1$ to the power $1/n$, which just gives $1$. Some people will write the imaginary unit $\mathbf i$ as $\sqrt{-1}$ (though most disapprove of this notation) even fewer would write $\mathbf i=(-1)^{\frac12}$, but hardly anybody would write $\mathbf i=1^{\frac14}$ or $\mathbf i\in1^{\frac14}$. – Marc van Leeuwen Nov 27 '16 at 19:02
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@MarcvanLeeuwen So, the mistake lies in notation, not in the reasoning, right? – vidyarthi Nov 27 '16 at 19:04
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Your mistake lies in interpreting exponential notation as being set-valued, while it is single valued. However the mistake of OP is either to introduce a non-integer power of $-1$ in the first place (which I would consider not defined at all), or for those who would admit this with some choice of definition, to rewrite $a^{p/q}$ as $\sqrt[q]{a^p}$ (with $a=-1$, $p=2$, $q=6$), which is definitely not valid when $a<0$. – Marc van Leeuwen Nov 27 '16 at 19:12
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@MarcvanLeeuwen So then, while solving equations like $x^{2/3}=c$ in the field $\mathbb{C}$, should we not write $x=c^{3/2}$? – vidyarthi Nov 27 '16 at 19:16
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Indeed you should not. But $x^{2/3}=c$ is not a proper equation in hte first place for $c\in\Bbb C\setminus\Bbb R$. One might write the equation $x^2=c^3$ if that is what you mean; it generally has $2$ solutions for $x$, neither of which I would write as $\sqrt{c^3}$, though some people might. – Marc van Leeuwen Nov 27 '16 at 19:18
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