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In the answer to this question llya has written that $y = x^\frac{1}{2}$ is defined as $y = ${$z: z^m = x$} so $y$ will have $m$ number of values(because $z^m$ is an equation of power $m$ so it has $m$ number of values. So $y$ has $m$ number of values.)

My question is that, is the statement given by llya correct? If it is then if $y = 1^\frac{1}{2}$ so $z^2 = 1$ therefore $z = 1, -1$ but the value of $1^\frac{1}{2}$ is not $-1$.

Ananth Kamath
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The Mathemagician
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  • $1^{\frac 12}=\sqrt 1=\pm 1$ – For the love of maths Aug 21 '18 at 17:06
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    Actually, nobody defines $x^{1/2}$ as ${z,;,z^2=x}$ and what you quote is a noncanonical try to extend the definition of the function $x\mapsto x^{1/2}$ from $[0,+\infty)$ to $\mathbb C$. What is interesting, from a sociological point of view, is that this version is periodically proposed on the site and that it receives some approval... – Did Aug 21 '18 at 17:08
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    @MohammadZuhairKhan The claim that $\sqrt{n^2}$ is $\pm n$ is raised often, too, but it destroys much of the usefulness of the $\sqrt{\phantom{0}}$ sign. Why not just write $\pm\sqrt u$ when you mean $\pm\sqrt u$? – David K Aug 21 '18 at 17:13
  • @DavidK I will keep that in mind. Thank you. – For the love of maths Aug 21 '18 at 17:15

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