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In our new books in freshman highschool,it is mentioned that :

When $a<0$ then, the expression

$$a^{m \over n} m,n\in \mathbb Z $$

even if the $n$ in the denominator is odd.

And my teacher gave the argument that:

$$((-8)^2)^{1 \over 6}=64^{1 \over 6}=2$$

But:$$(-8)^{2 \over 6} =(-8)^{1 \over 3}=-2$$

And he said it resulted into a contradiction that:$$ (a^m)^n=a^{mn}$$

My argument :we cannot make the step that:

$$(-8)^{2\over 6}=(-8)^{1\over3}$$ since there is no meaning ful way to go from the latter to the prior without changing the sign.

What is the correct answer if we take $a\in \mathbb C$or $a\in \mathbb R$?

miracle173
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Logan Luther
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  • this is the reason that we define $$\sqrt[n]{a}$$ only is $$a\geq 0$$ – Dr. Sonnhard Graubner Jan 14 '17 at 16:32
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    The sentence "When $a<0$ then, the expression $a^{m \over n}$ even if the $n$ in the is odd." is missing a few words, I think. – Arthur Jan 14 '17 at 16:33
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    @Dr.SonnhardGraubner We do define $\sqrt[n]a$ for odd $n$ and negative $a$. However, we only define fractional exponents $a^{1/n}$ for non-negative $a$, and thus $\sqrt[n]{a} = a^{1/n}$ only makes sense for non-negative $a$. – Arthur Jan 14 '17 at 16:35
  • @Arthur fixed.. – Logan Luther Jan 14 '17 at 16:35
  • @Arthur what if we only allow exponents that can be constructed from previous ones ,then will my argument in the post work? And depending on the starting point,we would only get one or the other of the answers. – Logan Luther Jan 14 '17 at 16:41
  • That's not the erroneous step. The argument fails because the rule $(x^a)^b = x^{ab}$ fails for $x<0$. See my my answer to this question http://math.stackexchange.com/questions/2086109/why-doesnt-xab-always-equal-xab/2086126#2086126 for an explanation. –  Jan 14 '17 at 16:42
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    @LoganLuther But we can make the step that $(-8)^{2/6} = (-8)^{1/3}$. The two numbers $\frac26$ and $\frac13$ are equal and must therefore be indistinguishable in any way. This includes that they should be indistinguishable as exponents. – Arthur Jan 14 '17 at 16:43
  • @OpenBall You can't say out of the blue that $(x^a)^b = x^{ab}$ fails for $x<0$. That depends entirely on your application and context. If, for instance, you work in number theory or combinatorics, the rule hods very well, many techniques and proofs depend on it, and you just have to accept that exponents can only be integers. – Arthur Jan 14 '17 at 16:44
  • @Arthur I would like to know more about this. Would you please suggest some reading? –  Jan 14 '17 at 16:45
  • I would say "open any book on elementary number theory or combinatorics". I assume it's in there, implicitly, all the time. I suppose you can agree, however, that at the very least, you can't have both $(x^a)^b = x^{ab}$ and fractional exponents for negative bases. I agree with you that in the context of analysis and complex numbers, one might want to take the route via the complex logarithm and define arbitrary powers for real numbers, but it's not a universal approach. – Arthur Jan 14 '17 at 16:47
  • @Arthur I read a book on elementary number theory and skimmed one on analytic number theory, but never encountered such a thing. If you could be more specific, e.g. give some link or suggest a book that you actually read, that would be great. –  Jan 14 '17 at 16:49
  • @can the downvoter explain his reasons? – Logan Luther Jan 14 '17 at 17:01

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If you want fractional exponents, then you have to accept that $\frac13 = \frac26$. Otherwise the exponents wouldn't deserve to be called fractions, and I for one do not know what rules such numbers would follow. This, along with the fat that we like the rule that $a^{mn} = (a^m)^n$, means that fractional exponents customarily only are defined for non-negative bases.

Arthur
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