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Let a be a real number and p a non zero real number. Then $a^p$ is also a real number. What definitions, propositions and etc are required to prove this?

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  • $\begingroup$ Related. $\endgroup$ – Cameron Buie Jun 16 at 22:41
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    $\begingroup$ I hope you’re specifying that $a$ be a positive real. Otherwise, you’re in serious trouble. $\endgroup$ – Lubin Jun 16 at 22:49
  • $\begingroup$ It’s pretty much meaningless for negative real numbers, isn’t it ? $\endgroup$ – Lubin Jun 16 at 23:27
  • $\begingroup$ @topologicalmagician a=-1, b=0.5? $\endgroup$ – piet.t Jun 17 at 7:49
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The answer to your question depends on where you want to begin. If you accept the real numbers and elementary calculus as given one approach is to define $e^x$ using appropriate tools - perhaps the power series. Then define the natural logarithm and, finally, $$ a^p = \exp(p \ln a) . $$ If you don't want to use the power series you can start by defining the natural logarithm as an integral, then the exponential function as its inverse.

This works for $a > 0$. For negative $a$ things get more complicated. The same final formula works, but the logarithm is not well defined. You need complex analysis to understand the way in which it is multivalued.

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Obviously, you must defined first the meaning of $a^p$, when $a,p\in\mathbb R$ and $p\neq0$. But I have never seen a definition which defines, say, $(-1)^{\sqrt2}$.

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  • $\begingroup$ Oh, I see. But is last expression you wrote real or not?? And why? May you please elaborate? $\endgroup$ – topologicalmagician Jun 16 at 22:30
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    $\begingroup$ As I wrote, it depends upon how you define it. $\endgroup$ – José Carlos Santos Jun 16 at 22:35

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