1. **State the problem:** We need to show that the function $f: \mathbb{R} \setminus \{0\} \to \mathbb{R} \setminus \{0\}$ defined by $f(x) = \frac{1}{x}$ is a bijection. 2. **Show that $f$ is injective (one-to-one):** Assume $f(x_1) = f(x_2)$ for some $x_1, x_2 \in \mathbb{R} \setminus \{0\}$. This means: $$\frac{1}{x_1} = \frac{1}{x_2}$$ Cross-multiplying gives: $$x_2 = x_1$$ Since $x_1 = x_2$, $f$ is injective. 3. **Show that $f$ is surjective (onto):** Take any $y \in \mathbb{R} \setminus \{0\}$. We want to find $x \in \mathbb{R} \setminus \{0\}$ such that: $$f(x) = y$$ That is: $$\frac{1}{x} = y$$ Solving for $x$: $$x = \frac{1}{y}$$ Since $y \neq 0$, $x = \frac{1}{y} \neq 0$, so $x \in \mathbb{R} \setminus \{0\}$. Thus, for every $y$ in the codomain, there exists an $x$ in the domain such that $f(x) = y$, so $f$ is surjective. 4. **Conclusion:** Since $f$ is both injective and surjective, it is a bijection. **Final answer:** $f(x) = \frac{1}{x}$ is a bijection from $\mathbb{R} \setminus \{0\}$ to $\mathbb{R} \setminus \{0\}$.