# Rational Exponent

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

• If whoever voted (and thought this is a reasonable question) can explain it in the comments, I'd be grateful. – Asaf Karagila Sep 21 '14 at 20:38
• @Asaf: This looks like a thoughtful question to me. How is $a^{p/q}$ defined for rational $p/q$ if $a$ is negative? A naive definition would indeed result in different values for $(-8)^{1/3}$ and $(-8)^{2/6}$. – TonyK Sep 21 '14 at 20:47
• @TonyK: I see. Thanks! – Asaf Karagila Sep 21 '14 at 20:49
• @TonyK, thx for the editing – Ivan Ehreshi Sep 21 '14 at 21:00

$1/3$ 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?
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|.
• I think your answer misses the point: What is $(-8)^{2/6}$? – TonyK Sep 21 '14 at 21:01
• But $(-8)^{1/3}$ is well-defined. – TonyK Sep 21 '14 at 21:59