# How to rewrite rational exponents where a negative value in format $a^\frac{m}{n}=(a^\frac{1}{n})^m=(a^m)^\frac{1}{n}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m$

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 $$a^\frac{m}{n}=(a^\frac{1}{n})^m=(a^m)^\frac{1}{n}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m$$?

• $a^{-b}=a^{-1\cdot b}=(a^{-1})^b = \frac1{a^b}$ – Don Thousand May 12 at 23:01
• 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. – Lubin May 13 at 0:38

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