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Well, I was studying laws of indices and the following steps came to my mind
$(-1)^{\frac {1}{2}}$
= $(-1)^{\frac {2}{4}}$
= $((-1)^{2})^{\frac {1}{4}}$
= $1^{\frac {1}{4}}$
= $1$
This is very strange, but I cannot find the mistake. In fact, $\frac {1}{2}$ and $\frac {2}{4}$ represent the same rational number and so the first step is alright. My second step is also correct as that is the definition. The remaining steps are also true. I'm confused!

Indrayudh Roy
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    Do not "Order" other users to "Find the mistake"... your last step $1^{\frac{1}{4}}=1$ shows that you are not yet familiar with roots of unity... there is no unique root of unity when you are out of the real numbers case.... If you are in real numbers then your first statement makes no sense so there is no wonder you last sentence makes no sense.. –  Dec 17 '13 at 13:57
  • See: http://math.stackexchange.com/questions/608023/what-is-1-frac23 (duplicate?) – Nigel Overmars Dec 17 '13 at 13:59

2 Answers2

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Within the rules of real number arithmetic, you start with nonsense since $(-1)^{1/2} = \sqrt{-1}$ is undefined, so you end with nonsense.

Within the rules of complex arithmetic, $\sqrt[4]{1}$ has 4 different values, $1, i, \pm \sqrt{i}$ ...

gt6989b
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You are probably thinking that $1/2=(1 \times 2)/(2\times 2) = 2/4$ and that is of course right, but when is an exponent what you are really doing is to square a number and taking the square root of it (or viceversa). The order is relevant! Or in other words $$ (-1)^{1/2}\neq ((-1)^2)^{1/4} $$ as you have correctly noticed. Or in otherwords taking the square of a number and then taking the square root are two distinct operations. Another example. Let's take $2$. Let's square it: $2^2=4$, now the square root of 4 is not only 2, but $\pm 2$. So you are not back to the original number. I hope is clearer now.

Umberto
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