8. A positive real number, its fractional part, integer part, and the number itself form a geometric sequence, then the fractional part of this positive real number is $\qquad$ .
8. $\frac{\sqrt{5}-1}{2}$.

Let the fractional part of this positive number be $a$, and the integer part be $b$.
From the problem, we know
$$
\begin{array}{l}
a(a+b)=b^{2} \Rightarrow a^{2}+a b-b^{2}=0 \\
\Rightarrow a=\frac{-1 \pm \sqrt{5}}{2} b .
\end{array}
$$

If $b=0$, then $a=0$, which is a contradiction.
Thus, $b>0, a>0$.
Therefore, $a=\frac{\sqrt{5}-1}{2} b$.
Since $a<1$, we have $b=1, a=\frac{\sqrt{5}-1}{2}$.
Hence, the fractional part of this positive number is $\frac{\sqrt{5}-1}{2}$.