Let $a_0$ be an irrational number such that $0 < a_0 < \frac 12$ . Define $a_n = \min \{2a_{n-1},1 - 2a_{n-1}\}$ for $n \geq 1$.
[list][b](a)[/b] Prove that $a_n < \frac{3}{16}$ for some $n$.
[b](b)[/b] Can it happen that $a_n > \frac{7}{40}$ for all $n$?[/list]
1. **Define the sequence and initial conditions:**
   Given \( a_0 \) is an irrational number such that \( 0 < a_0 < \frac{1}{2} \). Define the sequence \( \{a_n\} \) by:
   \[
   a_n = \min \{2a_{n-1}, 1 - 2a_{n-1}\} \quad \text{for} \quad n \geq 1
   \]

2. **Analyze the sequence:**
   Since \( a_0 \) is irrational and \( 0 < a_0 < \frac{1}{2} \), we can express \( a_0 \) in its binary expansion:
   \[
   a_0 = \sum_{k=2}^{\infty} b_k \cdot 2^{-k} \quad \text{where} \quad b_k \in \{0, 1\}
   \]

3. **Behavior of the sequence:**
   For \( n \geq 1 \), the sequence \( a_n \) is determined by the binary digits of \( a_0 \). Specifically:
   \[
   a_n = \begin{cases} 
   \sum_{k=2}^{\infty} b_{k+n} \cdot 2^{-k} & \text{if} \quad b_{n+1} = 1 \\
   \sum_{k=2}^{\infty} (1 - b_{k+n}) \cdot 2^{-k} & \text{if} \quad b_{n+1} = 0 
   \end{cases}
   \]

4. **Prove \( a_n < \frac{3}{16} \) for some \( n \):**
   - If there exist three consecutive zeros in the binary expansion of \( a_0 \), i.e., \( b_{n+1} = b_{n+2} = b_{n+3} = 0 \), then:
     \[
     a_n = \sum_{k=2}^{\infty} b_{k+n} \cdot 2^{-k} < \frac{1}{8}
     \]
   - If there exist three consecutive ones in the binary expansion of \( a_0 \), i.e., \( b_{n+1} = b_{n+2} = b_{n+3} = 1 \), then:
     \[
     a_n = \sum_{k=2}^{\infty} (1 - b_{k+n}) \cdot 2^{-k} < \frac{1}{8}
     \]
   - If neither of the above cases occurs, then \( a_0 \) would be rational, which contradicts the given condition that \( a_0 \) is irrational. Therefore, for an irrational \( a_0 \), there must exist some \( n \) such that \( a_n < \frac{3}{16} \).

5. **Can \( a_n > \frac{7}{40} \) for all \( n \)?**
   - Consider the binary representation of \( \frac{7}{40} \):
     \[
     \frac{7}{40} = 0.000111 \ldots_2
     \]
   - If \( a_n > \frac{7}{40} \) for all \( n \), then the binary expansion of \( a_0 \) must avoid patterns that would make \( a_n \) smaller than \( \frac{7}{40} \).
   - However, since \( a_0 \) is irrational, its binary expansion is non-repeating and must contain infinitely many zeros and ones. This implies that there will be some \( n \) such that \( a_n \) will be less than \( \frac{7}{40} \).

Conclusion:
\[
\boxed{\text{(a) Proved that } a_n < \frac{3}{16} \text{ for some } n.}
\]
\[
\boxed{\text{(b) It cannot happen that } a_n > \frac{7}{40} \text{ for all } n.}
\]