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 $(a, b)$.