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I know that by definition $x^a = \exp(a\log x)$ which is not defined when $x$ not in $\Re^∗_+$

Does it mean that for example $(-1)^3$ not defined for real numbers?

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  • $\begingroup$ There are two things that should be observed. First of all, $-1^3$ is to be interpreted as $-(1^3)=-1,$ so I suspect you mean to ask whether $(-1)^3$ is real. Second, since $(-1)^3:=-1\cdot-1\cdot-1,$ then yes, it is. The answer below explains the situation well. $\endgroup$ – Cameron Buie Oct 23 '15 at 23:17
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No, $(-1)^3= (-1)(-1)(-1)= -1$ is perfectly well defined. Your formula, $x^a= e^{a log(x)}$, is only correct when that logarithm exists.

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$(-1)^a$ is only real if $a$ is a rational number with an odd denominator. $$(-1)^a = \left\{ \begin{array}{rl} 1 & a=0 \\ -1& a\in\{ \frac{p}{2q+1},p\in Z_+ , q\in Z \} \\ undefined & otherwise\\ \end{array} \right.$$

which is a pretty bizarre discontinuous function

so $(-1)^3 = -1$

but $\lim_{x \to3}(-1)^x$ does not exist.

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