Question

Let \( S \) be the set of rational numbers with the following properties:
[I.] \( \frac{1}{2} \in S \)
[II.] If \( x \in S \), then both \( \frac{1}{x+1} \in S \) and \( \frac{x}{x+1} \in S \) Which of the following is true?

• A. \( S \) contains all rational numbers in the interval \( 0<x<1 \)
• B. \( S \) contains all rational numbers in the interval \( -1<x<1 \)
• C. \( S \) contains all rational numbers in the interval \( -1<x<0 \)
• D. \( S \) contains all rational numbers in the interval \( 1<x<\infty \)

Hint:

Track closure properties by applying the rules iteratively. Here, \( S \) expands across the interval \( 0<x<1 \) using the given operations.

Answer 1

The Correct Option is A

We are told that \( \frac{1}{2} \in S \), and the set is closed under the following operations: \[ \text{If } x \in S, \text{ then } \frac{1}{x+1} \in S \quad \text{and} \quad \frac{x}{x+1} \in S \] Let’s explore how the operations behave in the interval \( 0<x<1 \): Let \( x = \frac{1}{2} \). Then: 
\(\frac{1}{x+1} = \frac{1}{\frac{3}{2}} = \frac{2}{3} \in S\) 
\(\frac{x}{x+1} = \frac{\frac{1}{2}}{\frac{3}{2}} = \frac{1}{3} \in S\) Now repeat the process with \( \frac{2}{3} \): \[ \frac{1}{\frac{2}{3} + 1} = \frac{1}{\frac{5}{3}} = \frac{3}{5} \in S, \quad \frac{\frac{2}{3}}{\frac{5}{3}} = \frac{2}{5} \in S \] Similarly, for every \( 0<x<1 \), repeated applications of the transformations produce other values in that interval. Therefore, \( S \) includes all rational numbers in \( (0,1) \). 
\[ {S \text{ contains all rational numbers in } (0, 1)} \]