If Q is the set of rational numbers and P is the set of irrational numbers, then

Options

A. `P nn Q =phi `
B. `P sub Q`
C. `Q subP `
D. `P-Q =phi`

Answer

To solve the question, we need to analyze the sets \( P \) and \( Q \) and determine their intersection.

### Step-by-Step Solution:

1. **Define the Sets**:
- Let \( Q \) be the set of rational numbers. Rational numbers are numbers that can be expressed as the quotient of two integers, where the denominator is not zero. Examples include \( \frac{1}{2}, 3, -4, 0.75 \), etc.
- Let \( P \) be the set of irrational numbers. Irrational numbers cannot be expressed as a simple fraction. Examples include \( \sqrt{2}, \pi, e \), etc.

2. **Understand the Nature of the Sets**:
- Rational numbers (set \( Q \)) and irrational numbers (set \( P \)) are mutually exclusive. This means that no number can be both rational and irrational at the same time.

3. **Find the Intersection**:
- The intersection of two sets \( A \) and \( B \), denoted as \( A \cap B \), is the set of elements that are common to both \( A \) and \( B \).
- Since \( P \) and \( Q \) have no elements in common (as explained above), we can conclude that:
\[ P \cap Q = \emptyset \]
- In terms of cardinality, we can say that the intersection is equal to the empty set, which is denoted as \( \emptyset \).

4. **Conclusion**:
- Therefore, the intersection of the set of irrational numbers \( P \) and the set of rational numbers \( Q \) is empty:
\[ P \cap Q = \emptyset \]

### Final Answer:
The intersection of the set of rational numbers \( Q \) and the set of irrational numbers \( P \) is \( \emptyset \).

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