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There are several questions on the square root of a squared number. But mine is a little different, because under the pre-condition, that the square root function maps onto positive reals, using laws for radicals and exponents I do can verify for myself, that $\sqrt{x^2} = |x|$ and that $\left(\sqrt{x}\right)^2 = x$. However when I apply exponent laws I get a contradiction, therefore apparently exponent laws do not work for fractions as they do for whole numbers. However I have not heard of that distinction so far. Is this the case, or where is the fault in my reasoning?

An example:

\begin{align*} \left( \sqrt{-5} \right)^2 & = \left( \sqrt{5}i \right)^2\\ & = \left(\sqrt{5} \right)^2 (-1)\\ & = 5 (-1)\\ & = -5\\ \text{on the other hand:}\\ \sqrt{(-5)^2} & = \sqrt{25}\\ & = 5\\ \text{however:}\\ \left( \sqrt{-5} \right)^2 &= \left( (-5)^{\frac{1}{2}} \right)^2 = (-5)^1 = \left( (-5)^{2} \right)^{\frac{1}{2}} = \sqrt{(-5)^2} \end{align*}

Which would be a contradiction.

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  • $\begingroup$ under the pre-condition, that the square root function maps onto positive reals That's only true for $x \in \mathbb{R^+}\,$ and $-5$ is not a positive real number. $\endgroup$ – dxiv Jan 22 '17 at 20:57
  • $\begingroup$ @dxiv ok, but that does not resolve the problem as I see it, does it. $\endgroup$ – user3578468 Jan 22 '17 at 20:59
  • $\begingroup$ $(a^b)^c=a^{bc}$ safely when $a$ is positive (real or integer) and $b$ and $c$ are real, and also when $a$ is non-zero (real or complex) and $b$ and $c$ are integers. Your example does not fit either case $\endgroup$ – Henry Jan 22 '17 at 20:59
  • $\begingroup$ @user3578468 See for example What are the Laws of Rational Exponents?. $\endgroup$ – dxiv Jan 22 '17 at 21:01

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