Let f be a continuous real-valued function with domain (a, b). Show that if f(r) = 0 for each rational number r in (a, b), then f(x) = 0 for all x ∈ (a, b). Which statement is sccurate. 1) We will show that f(x) = 0 for all x in ( the real - the rationals). 2) We will show that f(x) = 0 for all x in ( the real - the rationals) intersect (a,b).

### Correct Statement:

The **accurate statement** is:

**2) We will show that \( f(x) = 0 \) for all \( x \in (\text{the real - the rationals}) \cap (a, b) \).**

### Explanation:

The proof leverages the fact that the rational numbers (\( \mathbb{Q} \)) are dense in the real numbers (\( \mathbb{R} \)), and \( f \) is continuous.

- The function \( f \) is given to satisfy \( f(r) = 0 \) for all rational \( r \in (a, b) \). 
- Since \( \mathbb{Q} \cap (a, b) \) is dense in \( (a, b) \), any real number \( x \in (a, b) \) can be approximated by a sequence of rational numbers \( \{r_n\} \) with \( r_n \to x \). 
- By continuity, \( f(x) = \lim_{n \to \infty} f(r_n) = 0 \), since \( f(r_n) = 0 \) for all \( r_n \).

The result shows \( f(x) = 0 \) for all \( x \in (a, b) \), which includes both rational and irrational numbers. Thus, **all points in the interval \( (a, b) \), including points outside the rationals, are shown to satisfy \( f(x) = 0 \).**

### Clarification on the Statements:

1. **Statement 1** suggests that \( f(x) = 0 \) for all \( x \in (\mathbb{R} \setminus \mathbb{Q}) \), which is not entirely accurate because the proof is specific to the interval \( (a, b) \).
2. **Statement 2** accurately limits the conclusion to \( (\mathbb{R} \setminus \mathbb{Q}) \cap (a, b) \), focusing on \( (a, b) \), which is the domain of \( f \).

Would you like further elaboration or help with the proof?  
---

#### Follow-Up Questions:
1. Why does the density of rationals play a crucial role in this proof?
2. How does the continuity of \( f \) guarantee the result for irrational points in \( (a, b) \)?
3. Can this result be extended to functions defined on \( \mathbb{R} \)?
4. What would happen if \( f \) were not continuous?
5. Is there a similar result for functions with domain \( [a, b] \) instead of \( (a, b) \)?

#### Tip:
Remember, continuity ensures that small changes in input result in small changes in output, which is why the behavior of \( f \) at rational points determines its behavior everywhere in \( (a, b) \).

Let f be a continuous real-valued function with domain (a, b). Show that if f(r) = 0 for each rational number r in (a, b), then f(x) = 0 for all x ∈ (a, b).

Explore the proof that if a continuous function f is zero at all rational numbers in an interval (a, b), it must also be zero for all real numbers in that interval. The solution leverages the density of rational numbers in real numbers and the properties of continuity. Suitable for advanced high school students studying calculus and real analysis.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation