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I'm covering rational exponents in my text book.

For all real numbers $a$ and natural numbers $m,n$, the following terms are equal (if they are all well-defined real numbers, some problems can occur otherwise): $a^\frac{m}{n}=(a^\frac{1}{n})^m=(a^m)^\frac{1}{n}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m$

The rule above makes sense to me somewhat and I've been able to follow along with some examples using these equivalent means of writing a rational exponent. For example, $8^\frac{2}{3}=(8^\frac{1}{3})^2=2^2=4$.

However, I was thrown of by the use of a negative rational exponent. How would the above equivalent way of writing a rational look with a negative rational exponent?

For example, $64^{-\frac{1}{3}}$? How would this appear in terms of $a^\frac{m}{n}=(a^\frac{1}{n})^m=(a^m)^\frac{1}{n}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m$?

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    $\begingroup$ $a^{-b}=a^{-1\cdot b}=(a^{-1})^b = \frac1{a^b}$ $\endgroup$ – Don Thousand May 12 at 23:01
  • $\begingroup$ To be precise, I point out that noninteger exponents are well-defined only when the base is a positive real number. If you try talking consistently about $(-4)^{1/6}$, you may find yourself in a quagmire. $\endgroup$ – Lubin May 13 at 0:38
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For a real number $a\neq0$, $n\geq0$, and $m>0$, $a^{-\frac n m}$ is defined as the multiplicative inverse of $a^{\frac n m}$ (all the other equations from your second line then apply); i.e. $$a^{-\frac n m}=\frac 1 {a^{\frac n m}}.$$

Thus, for example, $64^{-\frac13}=\dfrac1{64^{\frac13}}=\dfrac1{\sqrt[3]{64}}=\dfrac14$.

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