Amazing GP!

When $\{x\}$ , $\lfloor x \rfloor$ , and $x$ are in a geometric progression, then

$\begin{aligned} \{x\}\blue x & = \lfloor x \rfloor^2 & \small \blue{\text{Note that }x = \lfloor x \rfloor + \{x\}} \\ \{x\}\blue{(\lfloor x \rfloor + \{x\})} & = \lfloor x \rfloor^2 \\ \{x\}^2 + \lfloor x \rfloor \{x\} - \lfloor x \rfloor^2 & = 0 & \small \blue{\text{Solving the quadratic for }\{x\}} \end{aligned}$

$\begin{aligned} \implies \{x\} & = \frac {-\lfloor x \rfloor \blue + \sqrt{\lfloor x \rfloor ^2 + 4\lfloor x \rfloor ^2}}2 & \small \blue{\text{Since }\{x\} \ge 0} \\ & = \frac {(\sqrt 5 - 1)\lfloor x \rfloor}2 & \small \blue{\text{also }\{x\} < 1 \implies \lfloor x \rfloor = 1} \\ & = \frac {\sqrt 5 -1}2 = \blue \varphi - 1 & \small \blue{\text{where }\varphi = \frac {1+\sqrt 5}2 \text{ denotes the golden ratio.}} \end{aligned}$

Then $x = \lfloor x \rfloor + \{x\} = 1 + \varphi -1 = \varphi \approx \boxed{1.62}$ .

$\{x\}, \lfloor x\rfloor \text{ and } x \text{ are in a geometric progression, ratio} = r$

• $\textcolor{#3D99F6}{a = \{x\} } < 1 , \lfloor x\rfloor = a \cdot r \text{ is an integer, } x = a \cdot r^2$
• $\textcolor{#3D99F6}{ \{x\} \neq 0 } \hspace{20mm} \text{ else } r = \cfrac{\lfloor x\rfloor}{\{x\}} \text{is not well defined}$
• $\textcolor{#3D99F6}{ r = \cfrac{x}{\lfloor x\rfloor} > 1 } \hspace{15mm} \text{ as } x, \lfloor x\rfloor \text{ are same sign }$

$\begin{aligned} x - \lfloor x\rfloor - \{x\} &= 0 \\ \Rightarrow ar^2 - ar - a &= 0 \\ r^2 - r - 1 &= 0 \\ \implies r &= \cfrac{1 + \sqrt{5}}{2} \hspace{10mm} \text {neglect -ve as } r > 1 \end{aligned}$ $\begin{aligned} x - \lfloor x\rfloor &= \{x\} \\ ar^2 - ar &= a \\ \boxed{ar}(r-1) &= a \\ \boxed{ar}\left(\cfrac{\sqrt{5}-1}{2} \right) &= a \implies ar = 1 \hspace{10mm} \text{ as } \lfloor x\rfloor = ar \geq 2 \Rightarrow a > 1 \Rightarrow \text{contradiction} \\ \\ x &= ar^2 = ar\cdot r = 1 \cdot r \\ &= \boxed{ \cfrac{1 + \sqrt{5}}{2} } \hspace{20mm} \text{ which is } \phi \text{ golden ratio } \approx 1.618 \end{aligned}$

$\begin{aligned} \{x\} &= a = \cfrac{1}{r} \\ &= \cfrac{2}{1+\sqrt{5} } = \cfrac{2}{\sqrt{5}+1} \\ &=\cfrac{2}{\sqrt{5}+1} \cdot \cfrac{\sqrt{5}-1}{\sqrt{5}-1} \\ &= \cfrac{\sqrt{5}-1}{2} \approx 0.618 \end{aligned}$

We know that $x \{x\} = \lfloor x \rfloor ^2$ . Obviously $x$ cannot be negative, since the left-hand side (LHS) will be negative and the right-hand side (RHS) will be positive. So $x$ must be positive given the condition $x \ne 0$ . In addition, $x$ cannot be an integer, since the LHS will always be $0$ due to the term $\{x\}$ .

If $0 < x < 1$ , the RHS will be $0$ , and since $\{x\} \ne \lfloor x \rfloor \ne 0$ , there is no solution. If $2 < x$ however, then the RHS must be $4$ , which forces $\{x\} > \frac{4}{3}$ , but this is impossible. Hence $1 < x < 2$ . If $x = 1 + a$ , then we have the equation $a(1+a) = 1^2 \Rightarrow a^2+a-1 = 0$ .

Using the quadratic formula, we get that: $\boxed{a = \frac{1}{2}(\sqrt{5} - 1)}$

so $x = 1 + a = \frac{1 + \sqrt5}{2}$ , which is just the golden ratio. The other root must be discarded since it is negative.