My prof started today's lecture by writing $$(-2)^\frac{1}{4} = (-2)^{(2*\frac{1}{8})} = ((-2)^{2})^{\frac{1}{8}} = 4^{\frac{1}{8}}$$ and asked us whether this was valid or not. However he didn't provide any answer and just continued on from there.

In my opinion this is not valid because $(-2)^\frac{1}{4}$ should be a complex number whereas $4^{\frac{1}{8}}$ is a real number.

So my question is which step here is invalid? My guess would be the 2nd equality, and I'm guessing that $a^{mn}=(a^m)^n$ does not hold in every case. But I'm not entirely sure what the issue is here. Do we need $a>0$ ?


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    $\begingroup$ What does $(-2)^{1/4}$ mean ? $\endgroup$ – TheSilverDoe Sep 11 at 17:16
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    $\begingroup$ @JCAA not true, the solutions will just be complex $\endgroup$ – Riemann'sPointyNose Sep 11 at 17:28
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    $\begingroup$ @JCAA could you elaborate why you don't think so? Since ${(-2)^{\frac{1}{4}}}$ is usually defined as the solutions to the equation ${z^4=-2}$ which makes perfect sense $\endgroup$ – Riemann'sPointyNose Sep 11 at 17:33
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    $\begingroup$ @JCAA again, definable. Take the limit of the principle values of ${(-2)^{e_n}}$, where ${e_n}$ is any rational sequence converging to $e$. I'm not sure how that's relevant anyway, since OP isn't asking about irrational powers. It's rational. Just ask Wolfram alpha - it actually gives you these answers $\endgroup$ – Riemann'sPointyNose Sep 11 at 17:40
  • $\begingroup$ @JCAA Wolfram alpha disagrees $\endgroup$ – Riemann'sPointyNose Sep 11 at 17:46

First you need to define what you mean by expressions like ${(-2)^{\frac{1}{4}}}$. The usual way is to think of this "expression" as a set of solutions to the equation

$${z^4 = -2}$$

Likewise, by ${4^{\frac{1}{8}}}$ usually refers to the set of solutions for the equation

$${z^8 = 4}$$

In other words, ${4^{\frac{1}{8}}}$ isn't just a real number, and also will refer to some complex numbers that satisfy ${z^8 = 4}$.

If you go by this definition, you can see the second equation is actually just the first equation squared on both sides. If you square an equation on both sides like this - the new equation will contain solutions valid for both equations, but will also contain other dummy solutions that do not. In other words, solving ${z^8 = 4}$ will give solutions for both ${z^4 = -2}$ and ${z^4 = 2}$. Hence if you think of these expressions as sets, ${(-2)^{\frac{1}{4}}\neq 4^{\frac{1}{8}}}$, but ${(-2)^{\frac{1}{4}}\subset 4^{\frac{1}{8}}}$.

You have to be careful applying power rules to things involving non-integer powers pretty much. Hope that helps :)

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    $\begingroup$ This clears things up. Thank you! $\endgroup$ – Yakes Sep 11 at 18:41
  • $\begingroup$ @Yakes no problem! $\endgroup$ – Riemann'sPointyNose Sep 11 at 19:13

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