Problem 1709

1. Definitions and Assumptions:
- We call a positive integer $q$ a convenient denominator for a real number $\alpha$ if $\left| \alpha - \frac{p}{q} \right| < \frac{1}{10q}$ for some integer $p$.
- Assume $0 < \alpha, \beta < 1$ without loss of generality.
- Define a convenient denominator $q$ to be right-winged with respect to $r$ if $0 < \{qr\} < \frac{1}{10}$ and left-winged if $\frac{9}{10} < \{qr\} < 1$.

2. Convenient Denominators:
- A convenient denominator $q$ for $\alpha$ must be either left-winged or right-winged.
- We aim to show that if $\alpha$ and $\beta$ have the same set of convenient denominators, then either $\alpha + \beta$ or $\alpha - \beta$ is an integer.

3. Contradiction Setup:
- Assume there exist two convenient denominators $q_1$ and $q_2$ such that $q_1$ is left-winged and $q_2$ is right-winged with respect to $\alpha$, and both are similarly-winged with respect to $\beta$.
- Without loss of generality, assume $q_1$ and $q_2$ are both right-winged with respect to $\beta$.

4. Sequence Construction:
- Construct sequences $\{a_i\}$ and $\{b_i\}$ such that:
- $a_1 = 1$ and $b_1 = 0$,
- $(a_{i+1}, b_{i+1})$ is either $(a_i + 1, b_i)$ or $(a_i, b_i + 1)$,
- $a_i \cdot q_1 + b_i \cdot q_2$ is a convenient denominator.

5. Inductive Verification:
- Verify inductively that:
- If $0 < \{a_k \cdot q_1 + b_k \cdot q_2\} < \frac{1}{10}$, choose $a_{k+1} = a_k + 1$ and $b_{k+1} = b_k$.
- If $\{a_k \cdot q_1 + b_k \cdot q_2\} > \frac{9}{10}$, choose $(a_{k+1}, b_{k+1})$ to be the other possibility.

6. Contradiction:
- Since $a_N \cdot q_1 + b_N \cdot q_2$ is a convenient denominator of $\beta$, it implies:
$$\{ a_N \cdot q_1 \beta + b_N \cdot q_2 \beta \} < \frac{1}{10} \text{ or } > \frac{9}{10}$$
- However, analysis shows that once the value is greater than $\frac{1}{10}$, it must be less than $\frac{1}{5}$, leading to a contradiction.

7. **Elimination of $\alpha + \beta = 1$**:
- If $S_R(\alpha) = S_L(\beta)$, then $S_R(\alpha) = S_R(1 - \beta)$.
- Prove that $S_R(a) = S_R(b)$ implies $a = b$ for $0 < a, b < 1$.

8. Type-N Denominators:
- Divide convenient denominators into type-$N$ when they satisfy:
$$\frac{1}{10(N+1)} < \{q \alpha\} < \frac{1}{10N} \Leftrightarrow \frac{1}{10(N+1)} < \{q \beta\} < \frac{1}{10N}$$
- Prove that if $q$ is type-$M$ on $\beta$ and $M > N$, then $(N+1)q$ is not a convenient denominator of $\alpha$ while it is for $\beta$.

9. Final Contradiction:
- Assume $\{q \alpha\} \neq \{q \beta\}$.
- Let $\frac{1}{10} - \{q \alpha\} = r$ and $\frac{1}{10} - \{q \beta\} = s$ with $s > r$.
- Select $N$ large enough such that:
$$\frac{s}{r} > \frac{N+1+\frac{1}{r}}{N}$$
- Construct $C = q + D \cdot \left\lceil r(10(N+1)) \right\rceil$ and show it is not a convenient denominator of $\alpha$ but is for $\beta$.

10. Conclusion:
- The contradiction implies that either $\alpha + \beta$ or $\alpha - \beta$ must be an integer.

$\blacksquare$