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What wrong with this proof?

$(-1)=(-1)^{\frac{2}{2}}=(-1)^{2\times \frac{1}{2}}=\sqrt{1}=1$ then $1=-1$

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  • $\begingroup$ The issue is how you define $x \mapsto (x)^{\frac{1}{2}}$. $\endgroup$ – copper.hat Oct 31 '13 at 18:11
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$x^{\frac{1}{2}}$ is a multiple-valued "function", since in general $x$ has two square roots. One could also write:

$$\sqrt1=-1$$

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it is fake suppose f(x)=x and g(x)=(x)^(2/2)

f(x) is not equal to g(x) domain of both is R but g(x)=|x| instead of f(x)=x

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$(-1)^{2 \times\frac{1}{2}}$ is not equal to $\sqrt{1}$. The correct way to simplify $(-1)^{2 \times\frac{1}{2}}$, using universally accepted orders of operation (because they work), is to start by simplifying the exponent ${2 \times\frac{1}{2}}$ to $1$. Then you get $(-1)^1$ , which is equal to $-1$, which is where you started.

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