What is the value of $\sqrt{(-1)^2} ;$ 1 or -1?
$\sqrt{(-1)^2} = \sqrt 1= 1 $
$\sqrt{(-1)^2} = {((-1)^2)}^{1/2} = (-1)^1 = -1$
Or is it both?
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5By definition, $\sqrt x$ denotes the non-negative solution to $z^2=x$. Thus, for example, we always have $\sqrt {x^2}=|x|$, not $x$. – lulu Jan 11 '21 at 13:47
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It is the first. The equality ${((-1)^2)}^{1/2} = (-1)^1$ is incorrect. – TheSilverDoe Jan 11 '21 at 13:47
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@lulu for real $x$. – uniquesolution Jan 11 '21 at 13:48
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To @TheSilverDoe's point: there is a discrepancy between sqrt and $1/2$ power. These are not the same thing. – Cameron Williams Jan 11 '21 at 13:52
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If $z$ is a non-zero complex number and $m,n$ are positive integers prime to each other, then $(z^{1/n})^m=(z^m)^{1/n}$
If gcd$(m,n)\neq 1$ then $(z^{1/n})^m\neq (z^m)^{1/n}$, in general.
Example: Let $z=i,m=4,n=6.$ Then $(z^{1/n})^m=(i^{1/6})^4 \qquad \qquad \;\;=\cos {4r\pi+\pi \over 3}+i\sin {4r\pi+\pi \over 3}$ where $r=0,1,2,...,5$. It has three distinct values. But $(z^m)^{1/n}=1^{1/6}=\cos {2k\pi \over 6}+i\sin {2k\pi \over 6}$, where $k=0,1,...5$. It has six distinct values.
In your question $\sqrt {(-1)^2}=\sqrt 1=1$
[$\sqrt$ is the positive square root function].
Saikai Prime
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