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We define Square Roots as $$\sqrt{x^2} = \left|x\right| = \begin{cases} x, & \mbox{if }x \ge 0 \\ -x, & \mbox{if }x < 0. \end{cases}$$

However, if we take the Square Root of the Square Negative Number like $\sqrt{\left(-x\right)^{2}}$, a conflict arises.

For example, $\sqrt{\left(-5\right)^{2}}$ can be written as $\sqrt{25} = 5$.

Or we could write it as $\left(-5\right)^{2\cdot\frac{1}{2}} = \left(-5\right)^{1} = -5$. Both arguments seem logical to me.

I also thought about using the imaginary unit $i$. It gives me $\sqrt{5^{2}}\cdot i^{2} = -5$ which agrees with my second argument.

I am sorry if this question is too stupid. I can't make up my mind about which of those are correct.

J. W. Tanner
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eem
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2 Answers2

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We have no any conflict: $$\sqrt{(-x)^2}=\sqrt{x^2}=|x|.$$ $$\left((-5)^2\right)^{\frac{1}{2}}=25^{\frac{1}{2}}=5.$$ If we want to use the property $$a^{xy}=\left(a^x\right)^y$$ so we need $a>0$ by definition.

Michael Rozenberg
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You can't say $((-5)^2)^{\frac12}=(-5)^{2\cdot\frac12}. $ See this question and answers there, which explain that fractional powers of negative numbers are not uniquely defined, and the "rule" $(a^m)^n=a^{m\cdot n}$ does not always work when $m$ and $n$ are not integers.

J. W. Tanner
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