16. (16 points) Let $a, b \in \mathbf{R}$, and the function
$$
f(x)=a x^{2}+b(x+1)-2 \text {. }
$$

If for any real number $b$, the equation $f(x)=x$ has two distinct real roots, find the range of real numbers for $a$.
Three, 16. From the problem, we know the equation
$$
a x^{2}+(b-1) x+b-2=0
$$

has two distinct real roots. Therefore,
$$
\begin{array}{l}
\left\{\begin{array}{l}
a \neq 0, \\
\Delta=(b-1)^{2}-4 a(b-2)>0
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
a \neq 0, \\
b^{2}-2(1+2 a) b+8 a+1>0 .
\end{array}\right.
\end{array}
$$

For any real number $b$, equation (1) always holds, thus,
$$
\begin{array}{l}
\left\{\begin{array}{l}
a \neq 0, \\
\Delta=4(1+2 a)^{2}-4(8 a+1)<0
\end{array}\right. \\
\Rightarrow 0<a<1 .
\end{array}
$$