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Is $a^{p/q}$ equal to $a^{2p/2q}$? Do we need to simplify $p/q$ to its lowest terms? I need a strict mathematical definition which proves one or another statement. For example:

$(-8)^{1/3} = -2$

Is $(-8)^{2/6}$ equal to $\sqrt[6]{(-8)^2} = 2$?
Or $(\sqrt[6]{-8})^2$, which is undefined?
Or $(-8)^{1/3} =-2$?

TonyK
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Ivan Ehreshi
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2 Answers2

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$1/3$ and $2/6$ are the same number, so if we need to evaluate $(-8)^{1/3}$ and $(-8)^{2/6}$, we must make sure that the result is the same in each case. So yes, you need to reduce a fractional exponent to its lowest terms before applying it to a negative operand.

Having said that, the exponential function $x^y$ doesn't seem to be useful for negative $x$ and rational $y$, except perhaps when $y$ is of a special form to begin with, like $1/n$ for instance. So the question doesn't usually arise. Where did this question come from?

TonyK
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As long as $a\ge 0$, the function $a^x$ is well defined and therefore it doesn't matter how you represent x, either as $\frac 1 3$ or $\frac 2 6$, etc.

Once a hits negative numbers, you need to be careful because of the inability to take even powered roots of negative numbers, so for instance $(x^2)^\frac 1 2$ is not equal to x, but instead |x|.

Alan
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    I think your answer misses the point: What is $(-8)^{2/6}$? – TonyK Sep 21 '14 at 21:01
  • Here the point is that when you hit negative numbers, the operation is no longer well defined (as long as we are staying in the realm of real numbers and not the complex plane). So the answer (strictly in real numbers) depends on your order of operations. According to "standard" parsing, that would be -2, as you'd reduce the fraction first, but one could easily argue it's undefined. – Alan Sep 21 '14 at 21:53
  • But $(-8)^{1/3}$ is well-defined. – TonyK Sep 21 '14 at 21:59