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 $a\in \mathbb R$?

  • $\begingroup$ this is the reason that we define $$\sqrt[n]{a}$$ only is $$a\geq 0$$ $\endgroup$ – Dr. Sonnhard Graubner Jan 14 '17 at 16:32
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    $\begingroup$ 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. $\endgroup$ – Arthur Jan 14 '17 at 16:33
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    $\begingroup$ @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 $a$. $\endgroup$ – Arthur Jan 14 '17 at 16:35
  • $\begingroup$ @Arthur fixed.. $\endgroup$ – Logan Luther Jan 14 '17 at 16:35
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    $\begingroup$ @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. $\endgroup$ – Arthur Jan 14 '17 at 16:43

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.


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