We say that each positive number $x$ has two sons: $x+1$ and $\frac{x}{x+1}$. Characterize all the descendants of number $1$.
To characterize all the descendants of the number \(1\), we need to show that they are of the form \(\frac{F_{n\pm{1}}}{F_n}\), where \(n\) is an integer greater than \(1\) and \(F_n\) denotes the \(n\)th term in the Fibonacci sequence. We will use induction to prove this.

1. **Base Case:**
   - Start with \(x = 1\).
   - The Fibonacci sequence starts with \(F_1 = 1\) and \(F_2 = 1\).
   - For \(n = 2\), \(\frac{F_{2+1}}{F_2} = \frac{F_3}{F_2} = \frac{2}{1} = 2\).
   - For \(n = 2\), \(\frac{F_{2-1}}{F_2} = \frac{F_1}{F_2} = \frac{1}{1} = 1\).

2. **Inductive Step:**
   - Assume that for some \(n \geq 2\), the descendants of \(1\) are of the form \(\frac{F_{n\pm{1}}}{F_n}\).
   - We need to show that the descendants of \(\frac{F_{n+1}}{F_n}\) and \(\frac{F_{n-1}}{F_n}\) are also of the form \(\frac{F_{m\pm{1}}}{F_m}\) for some integer \(m\).

3. **Descendants of \(\frac{F_{n+1}}{F_n}\):**
   - The two sons of \(\frac{F_{n+1}}{F_n}\) are:
     1. \(\frac{F_{n+1}}{F_n} + 1 = \frac{F_{n+1} + F_n}{F_n} = \frac{F_{n+2}}{F_n}\).
     2. \(\frac{\frac{F_{n+1}}{F_n}}{\frac{F_{n+1}}{F_n} + 1} = \frac{\frac{F_{n+1}}{F_n}}{\frac{F_{n+1} + F_n}{F_n}} = \frac{F_{n+1}}{F_{n+1} + F_n} = \frac{F_{n+1}}{F_{n+2}}\).

4. **Descendants of \(\frac{F_{n-1}}{F_n}\):**
   - The two sons of \(\frac{F_{n-1}}{F_n}\) are:
     1. \(\frac{F_{n-1}}{F_n} + 1 = \frac{F_{n-1} + F_n}{F_n} = \frac{F_{n}}{F_n} = 1\).
     2. \(\frac{\frac{F_{n-1}}{F_n}}{\frac{F_{n-1}}{F_n} + 1} = \frac{\frac{F_{n-1}}{F_n}}{\frac{F_{n-1} + F_n}{F_n}} = \frac{F_{n-1}}{F_{n-1} + F_n} = \frac{F_{n-1}}{F_{n}}\).

5. **Conclusion:**
   - By induction, we have shown that all descendants of \(1\) are of the form \(\frac{F_{n\pm{1}}}{F_n}\), where \(n\) is an integer greater than \(1\).

\(\blacksquare\)

The final answer is \( \boxed{ \frac{F_{n\pm{1}}}{F_n} } \).