# Where is my mistake ?$(-1)^3=-1 \to \sqrt{-1}=-1$ [duplicate]

Where is my mistake ? (In the field of real numbers)

$$(-1)^3=-1 \to \sqrt{-1}=-1$$

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

$$-1=^?1$$

• Easier $-1=(-1)^{2/2}=\sqrt{(-1)^2}=1$. But also $(-1)^{2/2}=(\sqrt{-1})^2$... ooops. – A.Γ. Nov 26 '16 at 10:55
• This question has been asked here approximately $200$ times. – barak manos Nov 26 '16 at 10:59
• For 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

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 $1^\frac{1}{6} = \sqrt{1} = \pm 1$
So you get $-1=\pm 1$ and that is true.

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, $1,-1$

• 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
• @MarcvanLeeuwen But, arent there 2 sixth roots of unity in the field of reals? – vidyarthi Nov 27 '16 at 17:53
• 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
• @MarcvanLeeuwen So, the mistake lies in notation, not in the reasoning, right? – vidyarthi Nov 27 '16 at 19:04
• 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