115 Let $[x]$ denote the greatest integer not exceeding $x$, referred to as the integer part of $x$. Let $\{x\}=x-[x]$ denote the fractional part of $x$. If the integer part of $x$ is the mean proportional between $x$ and $\{x\}$, then the difference between $x$ and the reciprocal of $x$ is . $\qquad$
115 1. From the problem, we have
$$
x(x-[x])=[x]^{2}
$$

Rearranging gives
$$
x^{2}-[x] \cdot x-[x]^{2}=0 .
$$

Solving for $x$, we get
$$
x=\frac{1+\sqrt{5}}{2}[x],(x>0) .
$$

Since
$$
[x] \leqslant x<1+[x],
$$

it follows that
$$
[x] \leqslant \frac{1+\sqrt{5}}{2}[x]<1+[x] .
$$

Solving this, we get
$$
0 \leqslant[x]<\frac{\sqrt{5}+1}{2},
$$

Thus, $[x]=0,1$. When $[x]=0$, $x=0$, which has no reciprocal. Therefore, $[x]=1$, and we have
$$
x-\frac{1}{x}=\frac{1+\sqrt{5}}{2}-\frac{\sqrt{5}-1}{2}=1 \text {. }
$$