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Let suppose


if i take square root on both sides then


after multiplication and cancelation


which is wrong although i suppose a correct equation


where i am wrong and if i am correct then I just want to know why it is so


marked as duplicate by Zev Chonoles, Andrey Rekalo, Peter Taylor, A.S, Stefan Hansen Aug 9 '13 at 7:37

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  • 1
    $\begingroup$ You have run into a trick issue in mathematical analysis, which is that exponentiation with rational numbers can result in things that might seem unexpected to us - in this case, it's the fact that $(x^{2})^{\frac{1}{2}} \neq x$ for $x < 0$. To avoid ambiguity, in most real analysis texts (a branch of math that often answers questions like these), one DEFINES $\sqrt[n]{x}$ to be the POSITIVE number $y$ such that $y^{n} = x$. $\endgroup$ – Alex Wertheim Aug 9 '13 at 6:40

The square root of $(-1)^2$ is $1$, not $-1$. The identity $\sqrt{x^2} = x$ requires $x \geq 0$, and the general fact is $\sqrt{x^2} = |x|$.


You have to be careful.. $\sqrt{x^2} = |x|$ for all real numbers $x$. So $\sqrt{(-1)^2} = 1,$ not $-1$ as stated.


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