1. **Problem statement:** We have the function $$q(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{otherwise} \end{cases}$$ and we want to determine if $$q(x)$$ is continuous, and if the function $$xq(x)$$ is continuous, and if so, at which points. 2. **Recall the definition of continuity:** A function $$f$$ is continuous at a point $$a$$ if $$\lim_{x \to a} f(x) = f(a)$$. 3. **Analyze continuity of $$q(x)$$:** - For any real number $$a$$, consider sequences approaching $$a$$ from rationals and irrationals. - If $$x_n$$ is a sequence of rationals approaching $$a$$, then $$q(x_n) = 1$$. - If $$y_n$$ is a sequence of irrationals approaching $$a$$, then $$q(y_n) = 0$$. - Since these two limits differ, $$\lim_{x \to a} q(x)$$ does not exist. - Therefore, $$q(x)$$ is discontinuous at every point. 4. **Analyze continuity of $$xq(x)$$:** - Define $$f(x) = xq(x)$$. - For rational $$x$$, $$f(x) = x \cdot 1 = x$$. - For irrational $$x$$, $$f(x) = x \cdot 0 = 0$$. 5. **Check continuity at $$x=0$$:** - $$f(0) = 0q(0)$$. - Since $$0$$ is rational, $$q(0) = 1$$, so $$f(0) = 0 \cdot 1 = 0$$. - Consider $$\lim_{x \to 0} f(x)$$: - For rational $$x_n \to 0$$, $$f(x_n) = x_n \to 0$$. - For irrational $$y_n \to 0$$, $$f(y_n) = 0$$. - Both limits approach 0, so $$\lim_{x \to 0} f(x) = 0 = f(0)$$. - Hence, $$f(x)$$ is continuous at $$x=0$$. 6. **Check continuity at $$x = a \neq 0$$:** - $$f(a) = a q(a)$$. - If $$a$$ is rational, $$f(a) = a$$. - If $$a$$ is irrational, $$f(a) = 0$$. - Consider $$\lim_{x \to a} f(x)$$: - For rational $$x_n \to a$$, $$f(x_n) = x_n \to a$$. - For irrational $$y_n \to a$$, $$f(y_n) = 0$$. - Since $$a \neq 0$$, the two limits differ. - Therefore, $$f(x)$$ is discontinuous at every $$a \neq 0$$. **Final answers:** - $$q(x)$$ is discontinuous everywhere. - $$xq(x)$$ is continuous only at $$x=0$$ and discontinuous elsewhere.