1. **Stating the problem:** We want to prove the identity $$\left(\frac{a}{b}\right)^{-\frac{m}{n}} = \left(\frac{b}{a}\right)^{\frac{m}{n}}$$ where $a$, $b$, $m$, and $n$ are real numbers with $a \neq 0$, $b \neq 0$, and $n \neq 0$. 2. **Formula and rules used:** - Negative exponent rule: $$x^{-r} = \frac{1}{x^r}$$ for any nonzero $x$ and real number $r$. - Power of a fraction: $$\left(\frac{x}{y}\right)^r = \frac{x^r}{y^r}$$ for $y \neq 0$. 3. **Step-by-step proof:** $$\left(\frac{a}{b}\right)^{-\frac{m}{n}} = \frac{1}{\left(\frac{a}{b}\right)^{\frac{m}{n}}}$$ Using the power of a fraction rule: $$= \frac{1}{\frac{a^{\frac{m}{n}}}{b^{\frac{m}{n}}}}$$ Simplify the complex fraction: $$= \frac{b^{\frac{m}{n}}}{a^{\frac{m}{n}}}$$ Rewrite as a single fraction with positive exponent: $$= \left(\frac{b}{a}\right)^{\frac{m}{n}}$$ 4. **Explanation:** Raising a fraction to a negative exponent flips the fraction and changes the sign of the exponent to positive. This is why $$\left(\frac{a}{b}\right)^{-\frac{m}{n}}$$ equals $$\left(\frac{b}{a}\right)^{\frac{m}{n}}$$. **Final answer:** $$\boxed{\left(\frac{a}{b}\right)^{-\frac{m}{n}} = \left(\frac{b}{a}\right)^{\frac{m}{n}}}$$