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.
under the pre-condition, that the square root function maps onto positive realsThat's only true for $x \in \mathbb{R^+},$ and $-5$ is not a positive real number. – dxiv Jan 22 '17 at 20:57