1. **Problem statement:** We have a function $$f:\mathbb{R} \to \mathbb{R}$$ defined by $$ f(x) = \begin{cases} x & \text{if } x \text{ is rational} \\ 1 - x & \text{if } x \text{ is irrational} \end{cases} $$ We want to show that $$f$$ is continuous only at $$x = \frac{1}{2}$$. 2. **Recall the definition of continuity:** A function $$f$$ is continuous at a point $$c$$ if $$ \lim_{x \to c} f(x) = f(c). $$ This means the limit of $$f(x)$$ as $$x$$ approaches $$c$$ must exist and equal the function value at $$c$$. 3. **Evaluate $$f\left(\frac{1}{2}\right)$$:** Since $$\frac{1}{2}$$ is rational, $$ f\left(\frac{1}{2}\right) = \frac{1}{2}. $$ 4. **Check the limit at $$x = \frac{1}{2}$$:** For any sequence of rational numbers $$x_n \to \frac{1}{2}$$, $$ f(x_n) = x_n \to \frac{1}{2}. $$ For any sequence of irrational numbers $$y_n \to \frac{1}{2}$$, $$ f(y_n) = 1 - y_n \to 1 - \frac{1}{2} = \frac{1}{2}. $$ Since both rational and irrational sequences approaching $$\frac{1}{2}$$ give the same limit $$\frac{1}{2}$$, the limit exists and equals $$f\left(\frac{1}{2}\right)$$. 5. **Check continuity at any other point $$c \neq \frac{1}{2}$$:** - If $$c$$ is rational, then $$f(c) = c$$. - For rational sequences $$x_n \to c$$, $$f(x_n) = x_n \to c$$. - For irrational sequences $$y_n \to c$$, $$f(y_n) = 1 - y_n \to 1 - c$$. Since $$c \neq 1 - c$$ unless $$c = \frac{1}{2}$$, the limit does not exist (the two limits differ), so $$f$$ is not continuous at $$c \neq \frac{1}{2}$$. 6. **Conclusion:** The function $$f$$ is continuous only at $$x = \frac{1}{2}$$.