1
$\begingroup$

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.

$\endgroup$
5
  • 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

0
$\begingroup$

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 $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.