I'm doing a big fat calculus review, going through Paul Garrett's Calculus Refresher. So far it's very clear and concise, but I just got stuck at one point. He lays out a bit of review on exponents (what they define, how they combine, etc.), and then says:

$$a^{m/n} = \left(\sqrt[n]{a}\right)^{m}$$ One hazard is that, if we want to have only real numbers (as opposed to complex numbers) come up, then we should not try to take square roots, $4^{th}$ roots, $6^{th}$ roots, or any even order root of negative numbers.

Understood. No surprises there.

For general real exponents $x$ we likewise should not try to understand $a^x$ except for $a > 0$ or we’ll have to use complex numbers (which wouldn’t be so terrible). But the value of $a^x$ can only be defined as a limit: let $r_1, r_2, . . .$ be a sequence of rational numbers approaching $x$, and define $$a^x = \lim_{i}a^{r_i}$$ We would have to check that this definition does not accidentally depend upon the sequence approaching $x$ (it doesn’t), and that the same properties still work (they do).

...And with that, I'm lost. I think he's saying, first, that we'd be forced out of the comfort zone of real numbers if $a^x<0$ since we could end up with something like $-4^{1/2}$. But why can the value of $a^x$ only be defined as a limit? That part, and the sequence approaching $x$ just left me in the dust. Can anyone clarify this a bit? Thanks in advance!

very delayed update: As was pointed out in the comments below, it seems the limit comes into play if we want to define $a^x$ for all real values of $x$, including irrationals. This makes a lot more sense now. Thanks again.

  • 6
    $\begingroup$ Well, $a^n$ (where $n$ is an integer) is defined as repeated multiplication. The $n$-th root, or $a^{1/n}$, is defined by the inverse of this, and then all rational powers follow through $(a^{x})^{y}=a^{xy}$. But how do you define $a^{x}$ for arbitrary irrational $x$, except by imposing continuity? $\endgroup$
    – mjqxxxx
    Feb 19, 2013 at 6:33
  • $\begingroup$ Presumably some subscripts are missing and "let $r,r,\dots$ be a sequence" ought to be "let $r_1,r_2,\dots$ be a sequence". $\endgroup$ Feb 19, 2013 at 6:37
  • $\begingroup$ @mjqxxxx - I think I see what you're saying. Thanks for the help. $\endgroup$
    – ivan
    Feb 19, 2013 at 6:55
  • 1
    $\begingroup$ I would suggest rephrasing the title, as it does not describe the question. Also, @mjqxxxx - why not write that as an answer? $\endgroup$
    – yohBS
    Feb 19, 2013 at 7:20
  • $\begingroup$ @mjqxxxx, I think you nailed it. Thanks! I've been doing some algebra review lately, and just revisited this. Of course, it makes perfect sense now :) $\endgroup$
    – ivan
    Mar 21, 2013 at 19:24

1 Answer 1


If $ a > 0 $, then an easier way of defining $ a^{x} $ for any given $ x \in \mathbb{R} $ is $$ a^{x} \stackrel{\text{def}}{=} \exp(x \ln(a)). $$ As $ \exp $ is a continuous function on $ \mathbb{R} $, we have $$ \lim_{y \to x} \exp(y \ln(a)) = \exp \left( \lim_{y \to x} y \ln(a) \right) = \exp(x \ln(a)), $$ which implies that $$ \lim_{y \to x} a^{y} = a^{x}. $$ Therefore, we can also define $ a^{x} $ as the limit of the sequence $ (a^{q_{n}})_{n \in \mathbb{N}} $, where $ (q_{n})_{n \in \mathbb{N}} $ is any sequence in $ \mathbb{Q} $ that converges to $ x $.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .