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Let $\{x\}$ denote the fractional part of $x$. For example, $\{1.23\}=0.23$. I’m looking for positive real solutions to $\{x\}=\frac{1}{x}$. Right off the bat, I recognised that $\phi=\frac{1+\sqrt5}{2}$ is a solution, as $\phi-1=\frac{1}{\phi}$. I tried Wolfram Alpha, and it displays a few solutions before the page fully loads in, but after the page loads fully, the only solution it displays is $x\approx-12.0827625302982$. $\phi$ is one of the solutions Wolfram Alpha displays before the page loads in. Does anyone know what the other solutions are, and how to find them?

Heisenberg
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    Have you drawn a graph of both functions? There seem to be an infinity of such points. – MasB Jun 09 '22 at 14:25
  • I have, and I know that there’s probably an infinite number of solutions. Is there any way to find any algebraically, though? – Heisenberg Jun 09 '22 at 14:26
  • The solutions are continued fractions $[n:n,n,n,\cdots]$ as exposed in this interesting post https://math.stackexchange.com/q/1739576/399263 – zwim Jun 09 '22 at 14:56

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Simply from the graph of $\frac1x$ and $\{x\}$, it should be clear that, on every interval $(n, n+1]$ for $n\in\mathbb N$, there is exactly one solution to the equation $\frac1x = \{x\}$. *

Now, let's focus on one of those intervals. On that interval, we can write $x$ as $x=n + \{x\}$ for some $n$, then you see that in order for $\{x\} = \frac1x$ to hold, the following must hold:

$$\{x\} = \frac1x\\ x-n = \frac 1x\\ x^2-nx - 1 = 0.$$

* in fact, this statement can quite easily be shown to be true. For $n=1$, equality is achieved at $x=\phi$, as you showed. For $n\geq 2$, the equality is achieved somewhere inside the interval $[n, n+\frac12]$:

Let $f(x)=\frac1x$ and $g(x)=\{x\}$. Note that $f$ is a decreasing function on $(n,n+1)$, with $\frac{1}{n}>f(x)>\frac{1}{n+1}$ for all $x\in (n,n+1)$. Also, note that $g(n) = 0 < f(n)$, and $g(n+\frac12) = \frac12 > f(x)$. Also note that on $[n,n+\frac12]$, the functions $f$ and $g$ are continuous, so by the bisection theorem, they must intersect.

5xum
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  • I plotted the fractional part of the positive solution of the equation, and it’s inverse, and like you said, it holds for every natural number. Thanks a ton. – Heisenberg Jun 09 '22 at 14:31
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    It's worth adding that we want a root of $x^2-nx-1=0$ with $n \le x < n+1$. Since $x^2-nx-1$ is $-1$ when $x=n$ and $n$ when $x=n+1$, such a root must exist for all integers $n>0$ by the intermediate value theorem - formalizing the intuition we got from looking at the graph. – Misha Lavrov Jun 09 '22 at 14:33
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    @DanielSchepler True, so there is no solution on $[0,1]$. Edited. – 5xum Jun 09 '22 at 14:51
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    It might be worth stating explicitly that the solutions are of the form $\frac12(n+\sqrt{n^2+4})$ for integer $n\ge 1$, and that for large $n$ these are approximately $n+\frac1n-\frac1{n^3}$ – Henry Jun 09 '22 at 17:49