## Section48.3Degree of Accessibility of Even Fibonacci Numbers

Note: This is Theorem 3 in [48.7.2].

We will prove by contradiction, by providing a 3–colouring that has no monochromatic two terms with difference in $F_E\text{.}$

Here is some notation for the proof: For $x\in\mathbb{R}\text{,}$ $\{x\}$ is the fractional part of $x$ defined as $\{x\}=x-\lfloor x\rfloor\text{.}$

Next, the norm of $x$ is defined as, $\| x\|=\min(\{x \},1-\{x\}])\text{,}$ so $\| x\|$ is the shortest distance from $x$ to an integer. By Applying Binet's formula we get:

\begin{equation*} f_{n+1}-\varphi f_n= \frac{\varphi^{n+1}-(-\varphi)^{-n-1}}{\sqrt{5}}-\frac{\varphi^{n+1}+(-\varphi)^{-n+1}}{\sqrt{5}}\\ = -\frac{1}{\sqrt{5}}\left(\frac{1}{(-\varphi)^{n+1}}+\frac{1}{(-\varphi)^{n-1}}\right)\\ = \frac{\varphi^2+1}{\sqrt{5}\varphi}\frac{1}{(-\varphi)^n}= \frac{1}{(-\varphi)^n}\text{.}\\ \end{equation*}

Now we will show that $\lim_{n\to\infty}\|{\varphi\over2}f_{3n}=\frac{1}{2}\text{:}$

\begin{equation*} 0=\lim_{n\to\infty}\frac{1}{\varphi^{3n}}= \|f_{3n+1}-\varphi f_{3n}|\\ =\lim_{n\to\infty} \|\frac{1}{2}(f_{3n+1}-\varphi f_{3n})\|=\lim_{n\to\infty}\|\frac{1}{2}- {\varphi\over2} f_{3n})\|\text{,}\\ \end{equation*}

which implies $\lim_{n\to\infty}\|{\varphi\over2}f_{3n}\|= {1\over2}\text{.}$

Additionally $\frac{1}{3}\lt 0.382=\|{\varphi\over2}f_3\|\leq\|{\varphi\over2}f_{3n}\|\text{.}$

We define a $3$–colouring $\Psi:\mathbb{Z}^+\rightarrow\{1,2,3\}$ in the following way: For $C_1=[0,{1\over3})\text{,}$ $C_2=[{1\over3},{2\over3})\text{,}$ and $C_3=[{2\over3},1)\text{,}$ $\Psi(i)=j$ if and only if $\left\{\frac{\varphi i}{2}\right\}\in C_j\text{.}$

Now we show that that $F_E$ is not 3–accessible by letting $a$ and $f$ be integers with $f\in F_E\text{:}$

\begin{equation*} \|{1\over2}\varphi(a+f)-{1\over2}\varphi a\|=\|{1\over2}\varphi f\|>{1\over 3}\text{.} \end{equation*}

Thus $\Psi(a)\not= \Psi(a+f)$ because $a$ and $a+f$ are in distinct $C_i$'s.

Here is a drawing to explaining why $\{{\varphi\over2}a\}=A$ and $\{{\varphi\over2}(a+f)\}=B$ must lie on distinct $C_i$ intervals.

The black interval around $\{a\}$ is an inclusive neighbourhood around $\{a\}$ of length ${1\over3}\text{,}$ and $\{a+f\}$ cannot be in this neighbourhood.

We can now prove that $\text{doa }(F)\leq5$ by a $6$–colouring that is not accessible.

We define a colouring $\Phi:\mathbb{Z}^+\rightarrow\{1,2,3,4,5,6\}$ in the following way:

\begin{equation*} \Phi(x)=\left\{ \begin{array}{cl} \Psi(x)\amp \text{if } x \text{ is even}\\ \Psi(x)+3\amp \text{if } x \text{ is odd.}\\ \end{array} \right. \end{equation*}

Now $\Phi$–monochromatic terms must be an even distance apart, and by property of $\Psi$ there are no monochromatic terms that have a difference of $f\in F_E\text{.}$