Talk:Equivalence of Definitions of Real Exponential Function

If we want to keep $\exp x := e^x$ as a definition, then we have to clarify exactly what we're asserting. As it is, $e^x$ is defined as $\exp x \ln e$, which doesn't allow you to define \exp without being circular. Do we want to take the definition of integer and rational number indices and try to extend it to all real indices? Or scrap the def'n and just keep $e^x = \exp x \ln e$ as a justification of notation, rather than a definition? --GFauxPas 06:40, 8 February 2012 (EST)
 * They are now all interconnected except for the $e^x$ definition, which would be a nice, intellectually dishonest reason to scrap the $e^x$ definition. --GFauxPas 09:09, 8 February 2012 (EST)
 * Another suggestion: Keep the $\exp x = e^x$ as a definition for $x \in \Z, x \in \Q$, and prove the consistency for those. For $x \in \R \setminus \Q$, it will instead be a justification of notation. --GFauxPas 15:47, 9 February 2012 (EST)

There may be a theorem saying that a continuous function on $\Q$ (in the sense of the adapted $\epsilon$-$\delta$ definition) extends uniquely to an $\R$-continuous function. This will combine with the above for a proof of equivalence of definitions. That is, the $e^x$ notation well be denoting the unique extension of the function $q \to e^q$ on $\Q$. --Lord_Farin 17:19, 9 February 2012 (EST)
 * I took my last suggestion and implemented it. Please let us know if you find such a theorem. --GFauxPas 09:25, 10 February 2012 (EST)


 * A friend of mine immediately answered positively when I mentioned the theorem. It seems fairly straightforward; I haven't got a source, though. --Lord_Farin 18:03, 10 February 2012 (EST)