a) Prove that $\{x+y\}-\{y\}$ can only be equal to $\{x\}$ or $\{x\}-1$ for any $x,y\in \mathbb{R}$.

b) Let $\alpha\in \mathbb{R}\backslash \mathbb{Q}$. Denote $a_n=\{n\alpha\}$ for all $n\in \mathbb{N}^*$ and define the sequence $(x_n)_{n\ge 1}$ by

\[x_n=(a_2-a_1)(a_3-a_2)\cdot \ldots \cdot (a_{n+1}-a_n)\]

Prove that the sequence $(x_n)_{n\ge 1}$ is convergent and find it's limit.
### Part (a)

1. **Definition of fractional part**: Recall that for any real number \( x \), the fractional part \( \{x\} \) is defined as:
   \[
   \{x\} = x - \lfloor x \rfloor
   \]
   where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \).

2. **Expression for \( \{x + y\} \)**: Consider the expression \( \{x + y\} \). By definition:
   \[
   \{x + y\} = (x + y) - \lfloor x + y \rfloor
   \]

3. **Expression for \( \{x + y\} - \{y\} \)**: We need to evaluate \( \{x + y\} - \{y\} \). Using the definition of the fractional part:
   \[
   \{x + y\} - \{y\} = (x + y - \lfloor x + y \rfloor) - (y - \lfloor y \rfloor)
   \]
   Simplifying this, we get:
   \[
   \{x + y\} - \{y\} = x + y - \lfloor x + y \rfloor - y + \lfloor y \rfloor = x + (\lfloor y \rfloor - \lfloor x + y \rfloor)
   \]

4. **Possible values of \( \lfloor x + y \rfloor - \lfloor y \rfloor \)**: The term \( \lfloor x + y \rfloor - \lfloor y \rfloor \) can be either 0 or -1, depending on whether \( x + y \) crosses an integer boundary when added to \( y \). Specifically:
   - If \( \lfloor x + y \rfloor = \lfloor y \rfloor \), then \( \lfloor x + y \rfloor - \lfloor y \rfloor = 0 \).
   - If \( \lfloor x + y \rfloor = \lfloor y \rfloor + 1 \), then \( \lfloor x + y \rfloor - \lfloor y \rfloor = -1 \).

5. **Conclusion**: Therefore, the expression \( \{x + y\} - \{y\} \) can be either:
   \[
   \{x + y\} - \{y\} = x \quad \text{or} \quad \{x + y\} - \{y\} = x - 1
   \]
   Hence, \( \{x + y\} - \{y\} \) can only be equal to \( \{x\} \) or \( \{x\} - 1 \).

\(\blacksquare\)

### Part (b)

1. **Definition of \( a_n \)**: Given \( \alpha \in \mathbb{R} \setminus \mathbb{Q} \), define \( a_n = \{n\alpha\} \) for all \( n \in \mathbb{N}^* \).

2. **Expression for \( x_n \)**: The sequence \( (x_n)_{n \ge 1} \) is defined by:
   \[
   x_n = (a_2 - a_1)(a_3 - a_2) \cdots (a_{n+1} - a_n)
   \]

3. **Possible values of \( a_{n+1} - a_n \)**: Since \( \alpha \) is irrational, the sequence \( \{n\alpha\} \) is uniformly distributed in \([0, 1)\). Therefore, \( a_{n+1} - a_n \) can be either \( \alpha \) or \( \alpha - 1 \).

4. **Bounding \( |x_n| \)**: Let \( \beta = \max(\alpha, 1 - \alpha) \). Then:
   \[
   |a_{j+1} - a_j| \le \beta \quad \text{for all } j
   \]
   Hence:
   \[
   |x_n| = \left| (a_2 - a_1)(a_3 - a_2) \cdots (a_{n+1} - a_n) \right| \le \beta^n
   \]

5. **Limit of \( |x_n| \)**: Since \( \beta < 1 \), we have:
   \[
   \beta^n \to 0 \quad \text{as } n \to \infty
   \]
   By the squeeze theorem, \( |x_n| \to 0 \) as \( n \to \infty \).

6. **Conclusion**: Therefore, the sequence \( (x_n)_{n \ge 1} \) converges to 0.

\(\blacksquare\)

The final answer is \( \boxed{ 0 } \)