Question

Consider the function \[ f(x) = \begin{cases} 1, & \text{if } x \in (\mathbb{R} \setminus \mathbb{Q}) \cup \{0\}, \\ 1 - \frac{1}{p}, & \text{if } x = \frac{n}{p},\ n \in \mathbb{Z}\setminus\{0\},\ p \in \mathbb{N},\ \gcd(n,p)=1 . \end{cases} \] Then

• A. all \(x \in \mathbb{Q}\setminus\{0\}\) are strict local minima for \(f\).
• B. \(f\) is continuous at all \(x \in \mathbb{Q}\).
• C. \(f\) is not continuous at all \(x \in \mathbb{R}\setminus\mathbb{Q}\).
• D. \(f\) is not continuous at \(x = 0.\)

Hint:

Functions defined differently for rationals and irrationals (like Dirichlet-type functions) are typically discontinuous everywhere, but rational points can still form local extrema due to their isolated functional values.

Answer 1

The Correct Option is A

Step 1: Analyze the function.
For irrational \(x\), \(f(x) = 1.\) For rational \(x = \frac{n}{p}\) in lowest terms, \(f(x) = 1 - \frac{1}{p}<1.\)
Step 2: Continuity check.
Near any rational \(x_0 = \frac{n}{p}\), there are irrationals arbitrarily close with value \(1\). Thus, \(\lim_{x \to x_0} f(x) = 1 \ne f(x_0)\). Hence, \(f\) is discontinuous at every rational point.
Step 3: Local minima.
At rational points, \(f(x_0) = 1 - \frac{1}{p}\), and in any neighborhood of \(x_0\), \(f(x) \ge f(x_0)\) with strict inequality for nearby irrationals (\(f(x)=1\)). Hence, all rational points (except 0) are strict local minima.
Step 4: Conclusion.
Therefore, (A) is correct.