Question

Let \(S\) be the set of rational numbers with the following properties:
(i) \(\dfrac{1}{2} \in S\);
(ii) If \(x \in S\), then both \(\dfrac{1}{x+1} \in S\) and \(\dfrac{x}{x+1} \in S\).
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 \(1 < x < \alpha\)

Hint:

Check what interval both x/(x+1) and 1/(x+1) map (0,1) into, then try to reverse the moves starting from any rational in (0,1).

Answer 1

The Correct Option is A

Step 1: Understand the two moves.
We start with \(\frac{1}{2}\) in \(S\). From any \(x\) already in \(S\), rule (ii) lets us add two new numbers: \(\dfrac{1}{x+1}\) and \(\dfrac{x}{x+1}\).
First check: if \(0<x<1\), where do these two new numbers land?
\(x+1\) is between 1 and 2, so \(\dfrac{1}{x+1}\) is between \(\dfrac{1}{2}\) and \(1\).
Also \(\dfrac{x}{x+1} = 1 - \dfrac{1}{x+1}\), which is between \(0\) and \(\dfrac{1}{2}\).
So starting inside \((0,1)\), both moves keep us inside \((0,1)\): one move lands in the upper half \((\frac12,1)\), the other in the lower half \((0,\frac12)\). Since \(\frac12\in(0,1)\), by repeating the moves every number ever placed in \(S\) stays inside \((0,1)\), and is rational, since starting from a rational, both formulas only add and divide rationals, keeping the result rational. This already rules out options 2, 3 and 4, which all include numbers outside \((0,1)\) (negative numbers, or numbers at least 1).

Step 2: Show every rational in (0,1) can actually be reached, using the reverse moves.
Take any rational \(\dfrac{a}{b}\) in lowest terms with \(0<a<b\). Run the two forward moves backward:
if \(y=\dfrac{1}{x+1}\) then \(x=\dfrac{1-y}{y} = \dfrac{b-a}{a}\);
if \(y=\dfrac{x}{x+1}\) then \(x=\dfrac{y}{1-y} = \dfrac{a}{b-a}\).

Step 3: Apply the correct backward move depending on whether a/b is above or below 1/2.
If \(\dfrac{a}{b} > \dfrac12\) (that is \(2a>b\)), it must have come from the first move, so go back to \(\dfrac{b-a}{a}\); here the new denominator is \(a\), smaller than \(b\).
If \(\dfrac{a}{b} < \dfrac12\) (that is \(2a<b\)), it must have come from the second move, so go back to \(\dfrac{a}{b-a}\); here the new denominator is \(b-a\), smaller than \(b\).
Either way, the denominator strictly shrinks, exactly like a step of the Euclidean algorithm on \(a\) and \(b\), which are coprime.

Step 4: Conclude the process must end at 1/2.
Since the denominator keeps strictly shrinking at each backward step, this process cannot continue forever, so it must terminate. The only place it can terminate, given rule (i), is at \(\dfrac{1}{2}\), the starting point of \(S\). Reversing all these steps forward shows \(\dfrac{a}{b}\) is reachable from \(\frac12\), so it belongs to \(S\).
Since \(\dfrac{a}{b}\) was any rational in \((0,1)\), every rational number in \((0,1)\) belongs to \(S\). Combined with Step 1, \(S\) is exactly the set of rationals in \((0,1)\).

Final Answer:
Option 1 is correct: S contains all rational numbers in the interval \(0<x<1\).
\[ \boxed{0 < x < 1} \]