10. A positive number, if its fractional part, integer part, and the number itself form a geometric sequence, then the number is
Let the positive number be $x$, and its integer part be $[x]$, then $[x]^{2}=x(x-[x])$. Clearly, $x$ is not an integer, and $0 < x - [x] < 1$. When $[x] > 1$, $x(x-[x])=[x]^{2}>(x-1)^{2} \Rightarrow (2-[x]) x > 1 \Rightarrow [x]=1$. Thus, $x^{2}-x-1=0 \Rightarrow x=\frac{1+\sqrt{5}}{2}$, which satisfies the condition upon verification.