Example 9 Let $f(x)$ be a monotonic continuous function defined on the set of real numbers $\mathrm{R}$. Solve the functional equation $f(x) \cdot f(y)=f(x+y)$.
Solution: From $f(x) \cdot f(y)=f(x+y)$, using induction we get: $f\left(x_{1}\right) f\left(x_{2}\right) \cdots f\left(x_{n}\right)=f\left(x_{1}+x_{2}+\cdots x_{n}\right)$,
when $x_{1}=x_{2}=\cdots=x_{n}$,
we have $[f(x)]^{n}=f(n x)$.
If $x=1, f(n)=[f(1)]^{n}$, let $f(1)=a$, then $f(n)=a^{n}$.
In equation (1), let $x=\frac{1}{n}$ to get: $\left[f\left(\frac{1}{n}\right)\right]^{n}=f(1)$.
Since $f(x)$ is defined on the real number set $\mathbf{R}$, when $n$ is even, it must be that $f(1) \geqslant 0$, thus $a \geqslant 0, \therefore f\left(\frac{1}{n}\right)=a^{\frac{1}{n}}$.
If $m$ is a positive integer, using the above formula we get:
$$
\begin{array}{l}
f\left(\frac{m}{n}\right)=f\left(m \cdot \frac{1}{n}\right)= \\
f\left(\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}\right)=\left[f\left(\frac{1}{n}\right)\right]^{m}= \\
\left(a^{\frac{1}{n}}\right)^{m}=a^{\frac{m}{n}} .
\end{array}
$$

In the original equation, let $y=0$ to get: $f(x) \cdot f(0)=$ $f(x)$, since $f(x)$ is monotonic and $f(x)$ is not always 0,
$$
\therefore f(0)=1=a^{0} \text {. }
$$

In the original equation, let $y=-x$, we have $y=-x=-\frac{m}{n}$ $(n, m \in \mathbf{N})$, then $f\left(-\frac{m}{n}\right) \cdot f\left(\frac{m}{n}\right)=f(0)$, i.e., $f\left(-\frac{m}{n}\right)=\frac{1}{f\left(\frac{m}{n}\right)}=\left(a^{\frac{m}{n}}\right)^{-1}=a^{-\frac{m}{n}}$ (since $f\left(-\frac{m}{n}\right)$ is meaningful, $\left.\therefore a>0\right)$.

Thus, we have found the function equation $f(x)=a^{x}(a>0)$ within the set of rational numbers,

and since $f(x)$ is monotonic, it cannot always be 1, so $f(x)=a^{x}$ $(a>0$ and $a \neq 1)$ is an exponential function.

When $a=\alpha$ is an irrational number, let $a_{i}0$ and $a \neq 1)$.