If $x$, $\{x\}$, $\lfloor x\rfloor$ are in G.P, find $x$.

Question

If $x$, $\{x\}$, $\lfloor x\rfloor$ are in Geometric Progression, find $x$; $x \neq 0$.

Here, $\{x\}=x-\lfloor x\rfloor$

Some properties are pretty evident: $$0\leq \{x\} < 1 \tag{1}$$ $$\lfloor x \rfloor \leq x\tag{2}$$

But I can't seem to be able to use them to substitute and find an equation for $x$. Any help is appreciated.

Answer 1

Basic Approach. We assume that $x > 1$. (This is a mistake; Michael's answer gives a solution where $x < 0$.)

In this case, let the three numbers be denoted, instead, $r, n, n+r$, where $n$ is a positive integer and $r$ is a real value such that $0 \leq r < 1$. We can quickly determine that $r > 1/2$ (why?) and so $n = 1$ (why?). Then find $r$ such that

$$ r = \frac{1}{1+r} $$

You should obtain a reasonably familiar number (for $1+r$).

Answer 2

Let $x=n+y$, $n=\lfloor x\rfloor$ and $y=\{x\}$ Since these numbers are in geometric progression, we have $x/y=y/n$, or

$$\frac {n+y}{y}=\frac yn$$ $$(n+y)n=y^2$$ $$y^2 - n^2 - ny = 0$$ $$y=\frac{n\pm n\sqrt{5}}2$$

$y$ is between 0 and 1, so $n(1\pm\sqrt{5})/2$ has to be between $0$ and $1$.
This is impossible for $(1+\sqrt{5})/2$ as the first positive multiple of 1.618 is more than 1. But for $(1-\sqrt{5})/2\approx-0.618$, we have have $n=-1$

So $$n=-1\\y=n\frac{1-\sqrt{5}}2=\frac{\sqrt{5}-1}2$$ $$x=n+y=\frac{\sqrt{5}-3}2 $$

So $x=-\phi^{-2}$ and $\{x\}=\phi^{-1}$