I've been reading through a course on exponential functions, starting from integer-valued exponents to rational ones as in: $x^r$ from $r\in \Bbb{N}$ to $\Bbb{Z}$, and combining them to rigorously construct for $r\in\ \Bbb{Q}$. Still, this book adresses high-schoolers, and therefore "summons" an extension of the exponential notation for real exponents. Is there any formal basis for this extension? Could it be related to the density of $\Bbb{Q}$ in $\Bbb{R}$ ? Or the limit of a sequence $(x^{(r_n)})$ where $(r_n)_{n=1}$ is a sequence of rationals that converge to $r$ a non-rational exponent?


The answer is: yes, it is related to both those things. Given any real $\alpha,$ the density of $\Bbb Q$ in $\Bbb R$ allows us to find an increasing sequence $(r_n)_{n=1}^\infty$ of rational numbers converging to $\alpha$. Depending on which positive $x$ we are considering, the sequence $(x^{r_n})_{n=1}^\infty$ will be increasing, decreasing, or constant. In the last case, the sequence readily converges, and its limit is defined to be $x^\alpha.$ In the first two cases, we note that the sequence is bounded, so by completeness of $\Bbb R$, has a least upper/greatest lower bound, one of which will be the limit of the sequence, and in either case, we define $x^\alpha$ to be the limit of the sequence.

Showing that our choice of sequence doesn't matter (so that $x^\alpha$ is well-defined in this fashion) is a bit trickier, and likely beyond the scope of your course, but that's the idea.

  • $\begingroup$ Choice-of-sequence can actually be handled pretty cleanly, I think, with a Dedekind cut construction - the key (and admittedly the tricky part of the construction) is that for any given $x\gt 0$, $x^q$ is monotonic (either monotonic up or monotonic down), but aside from the messiness of a quick case-based analysis (splitting $x\gt 1$ from $x\lt 1$) the only thing that's used is 'clearing denominators', comparing $x^{r/s}$ with $x^{p/q}$ by raising both to the power $sq$. $\endgroup$ – Steven Stadnicki Nov 26 '13 at 18:36

Yes, it is related to both of the things that you mention. Because $\mathbb{Q}$ is dense in $\mathbb{R}$, a continuous function on $\mathbb{Q}$ (for example, the function given by $q \mapsto x^q$ where $x > 0$) can have at most one continuous extension to $\mathbb{R}$. This is because density implies that every real number $r$ is a limit of a sequence of rationals $\{q_n : n \in \mathbb{N}\}$, so any continuous extension must satisfy $x^r = \lim_{n \to \infty} x^{q_n}$. To see that a continuous extension exists at all, we must check that $\lim_{n \to \infty} x^{q_n}$ exists and is indpendent of the choice of sequence $\{q_n : n \in \mathbb{N}\}$ converging to $r$.

Alternatively, a simpler definition of $x^r$ would be $$x^r = \begin{cases} \sup \{x^q : q \in \mathbb{Q} \mathbin{\&} q \le r\} & \text{if } x \ge 1,\\ \sup \{x^q : q \in \mathbb{Q} \mathbin{\&} q \ge r\} & \text{if } 0 < x < 1, \end{cases} $$ although it may take some work to check that this defines a function with the desired properties.

  • $\begingroup$ @CameronBuie Thanks, I will fix it. $\endgroup$ – Trevor Wilson Nov 26 '13 at 18:34
  • $\begingroup$ No problem! +1 for providing a simpler alternative. $\endgroup$ – Cameron Buie Nov 26 '13 at 18:38

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