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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Related. – Cameron Buie Jun 16 '19 at 22:41
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4I hope you’re specifying that $a$ be a positive real. Otherwise, you’re in serious trouble. – Lubin Jun 16 '19 at 22:49
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It’s pretty much meaningless for negative real numbers, isn’t it ? – Lubin Jun 16 '19 at 23:27
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@topologicalmagician a=-1, b=0.5? – piet.t Jun 17 '19 at 07:49
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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.
Ethan Bolker
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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, $(-1)^{\sqrt2}$.
José Carlos Santos
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Oh, I see. But is last expression you wrote real or not?? And why? May you please elaborate? – Jun 16 '19 at 22:30
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