0
$\begingroup$

$(-3)^{\frac{3}{2}}=-3\sqrt{3}i$

$(-3)^{\frac{6}{4}}=\sqrt{27}$

(not the same thing).

What's the deal? It's interesting because people work with fractional exponents all the time and I've never seen someone bother to check whether the top and bottom maintain their parity when canceling, but clearly it makes a difference if you can have a negative base.

More precisely, how are exponents (especially of negative numbers) defined (in a rigorous sense), so that I can understand the problem here?

I don't think the solution is just to form a convention in which you simplify as much as possible before doing operations. I know it would give consistent results, but by the same reasoning, we could have chosen 0!=0. We chose not to make it that way for good reason. There are many applications in which 0! = 1 is the only elegant possibility. Having asked that... does anyone know of applications of this sort of thing?

$\endgroup$
  • $\begingroup$ related $\endgroup$ – Ilya Nov 25 '14 at 9:59
  • $\begingroup$ Looks almost identical to a question I posted a few days ago: math.stackexchange.com/q/1030633/131263. $\endgroup$ – barak manos Nov 25 '14 at 10:03
  • $\begingroup$ Ah, didn't see that when I searched. But I have a few more questions in this post that weren't asked in yours, so I won't delete. $\endgroup$ – doublefelix Nov 25 '14 at 10:12
  • $\begingroup$ Didn't mean you should delete, but some of the answers to that question might help with yours... $\endgroup$ – barak manos Nov 25 '14 at 10:27
1
$\begingroup$

You have to be careful with these kind of things if your base is not a non-negative real number. For example, $$1=1^{1/2}=[(-1)^2]^{1/2}=-1.$$ The reason for this can be found when you look at the "true" definition of $x\mapsto a^x$ when $a\in\mathbb{C}\setminus[0,\infty)$. We define this as: $$a^x:=e^{x\log a}$$ which of course requires some kind of definition for the logarithm. The usual one is $$\log x:=\log|x|+i\arg x$$ which is of course multi-valued since $\arg$ is. We thus don't have, in general, $(ab)^c=a^cb^c$ or $(a^b)^c=a^{bc}$.

$\endgroup$

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.