Let `f: [0,3] rarr R` be defined by `f(x)= "min" {x-[x], 1+[x]-x}` where [x] is the greatest integer less than or equal to x. Let P denote the set containing all `x in [0,3]` where f is discontinuous, and Q denote the set containing all `x in (0,3)` where f is not differentiable. Then the sum of number of elements in P and Q is equal to_____

Answer

To solve the problem, we need to analyze the function \( f(x) = \min \{ x - [x], 1 + [x] - x \} \) where \( [x] \) is the greatest integer less than or equal to \( x \). The goal is to determine the sets \( P \) and \( Q \) and find the sum of their cardinalities.

### Step-by-Step Solution:

1. **Understanding the Function**:
- The function \( f(x) \) is defined as the minimum of two expressions: \( x - [x] \) (the fractional part of \( x \)) and \( 1 + [x] - x \).
- The fractional part \( x - [x] \) is continuous and varies between 0 and 1 as \( x \) increases from an integer to the next integer.
- The second part \( 1 + [x] - x \) can be rewritten as \( 1 - (x - [x]) \). This also varies between 0 and 1.

2. **Identifying Points of Discontinuity (Set \( P \))**:
- The function \( f(x) \) can potentially be discontinuous at integer points where the behavior of the fractional part changes.
- Since both \( x - [x] \) and \( 1 + [x] - x \) are continuous functions, we need to check if they are equal at integer points.
- At integers \( x = 0, 1, 2, 3 \):
- \( f(0) = \min\{0, 1\} = 0 \)
- \( f(1) = \min\{0, 1\} = 0 \)
- \( f(2) = \min\{0, 1\} = 0 \)
- \( f(3) = \min\{0, 1\} = 0 \)
- Since \( f(x) \) does not change value at these points, \( f(x) \) is continuous everywhere in the interval [0, 3]. Thus, \( P = \emptyset \) and \( |P| = 0 \).

3. **Identifying Points of Non-Differentiability (Set \( Q \))**:
- The function \( f(x) \) can be non-differentiable at points where the two expressions switch dominance.
- The switch occurs at points where \( x - [x] = 1 + [x] - x \).
- Solving \( x - [x] = 1 + [x] - x \) gives \( 2x = 1 + 2[x] \) or \( x = \frac{1 + 2[x]}{2} \).
- This occurs at \( x = 0.5, 1.5, 2.5 \) (the midpoints between integers) and also at the integers \( 1, 2 \) where the function value changes.
- Therefore, the points of non-differentiability are \( x = 0, 1, 1.5, 2, 2.5, 3 \).
- Thus, \( |Q| = 5 \).

4. **Calculating the Final Result**:
- The sum of the number of elements in \( P \) and \( Q \) is:
\[ |P| + |Q| = 0 + 5 = 5 \]

### Final Answer:
The sum of the number of elements in \( P \) and \( Q \) is \( \boxed{5} \).