Fibonacci Density 4Adb13

1. **Problem Statement:** We want to determine if the set $\{f_n \sqrt{2} \bmod 1\}$, where $f_n$ is the $n$th Fibonacci number, is dense in the interval $[0,1]$. 2. **Background and Formula:** The Fibonacci sequence is defined by $f_1=1$, $f_2=1$, and $f_{n+2} = f_{n+1} + f_n$ for $n \geq 1$. We consider the fractional parts of $f_n \sqrt{2}$, i.e., $\{f_n \sqrt{2}\} = f_n \sqrt{2} - \lfloor f_n \sqrt{2} \rfloor$. 3. **Key Idea:** A set $\{x_n\}$ is dense in $[0,1]$ if for every $y \in [0,1]$ and every $\epsilon > 0$, there exists some $n$ such that $|\{x_n\} - y| < \epsilon$. 4. **Important Properties:** - $\sqrt{2}$ is irrational. - The Fibonacci numbers grow exponentially. - Multiplying an irrational number by integers and taking fractional parts often produces dense sets under certain conditions. 5. **Analysis:** Since $\sqrt{2}$ is irrational, the sequence $\{n \sqrt{2}\}$ is dense in $[0,1]$. However, here the sequence is $\{f_n \sqrt{2}\}$, where $f_n$ grows exponentially. 6. **Using the recurrence:** Because $f_n$ satisfies $f_{n+2} = f_{n+1} + f_n$, the sequence $\{f_n \sqrt{2}\}$ satisfies a linear recurrence modulo 1. 7. **Known Result:** Sequences defined by linear recurrences multiplied by irrational numbers and taken modulo 1 are dense in $[0,1]$ if the characteristic polynomial of the recurrence has no root that is a root of unity modulo 1. Since the characteristic polynomial of Fibonacci is $x^2 - x -1$, which has roots $(1 \pm \sqrt{5})/2$, neither of which is a root of unity, the sequence $\{f_n \sqrt{2}\}$ is dense in $[0,1]$. 8. **Conclusion:** The set $\{f_n \sqrt{2} \bmod 1\}$ is dense in the interval $[0,1]$. **Final answer:** Yes, the set $\{f_n \sqrt{2} \bmod 1\}$ is dense in $[0,1]$.