5. Let $[a]$ denote the greatest integer not exceeding the real number $a$, and let $\{a\}=a-[a]\left(a \in \mathbf{R}_{+}\right)$. If $a$, $[a]$, and $\{a\}$ form a geometric sequence in order, then $a=$ $\qquad$

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5. $\frac{1+\sqrt{5}}{2}$.

Let $\{a\}=\theta(\theta \in[0,1)),[a]=k$.
By the problem, $\theta>0, k>1$, and $a\{a\}=[a]^{2}$
$$
\begin{array}{l}
\Rightarrow(\theta+k) \theta=k^{2} \\
\Rightarrow \theta^{2}+k \theta-k^{2}=0 \\
\Rightarrow \theta=\frac{-1+\sqrt{5}}{2} k \text { (negative value discarded). }
\end{array}
$$

Since $0<\theta<1, k$ is a positive integer, thus, $k=1$.
Therefore, $\theta=\frac{\sqrt{5}-1}{2}$.
Hence, $a=k+\theta=\frac{1+\sqrt{5}}{2}$.