Example 1 Let $f(x)$ be a monotonic function defined on the set of real numbers $\mathbf{R}$. Solve the functional equation $f(x) \cdot f(y)=f(x+y)$.
Solve: From $f(x) \cdot f(y)=f(x+y)$, using mathematical induction, we get
$f\left(x_{1}\right) \cdot f\left(x_{2}\right) \cdots f\left(x_{n}\right)=f\left(x_{1}+x_{2}+\cdots+x_{n}\right)$.
When $x_{1}=x_{2}=\cdots=x_{n}=x$, 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, let $x=\frac{1}{n}$, we 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 $f(1) \geqslant 0$, so $a \geqslant 0$. In this case, $f\left(\frac{1}{n}\right)=$ if $m$ is a positive integer, using the above equation, we get
$$
f\left(\frac{m}{n}\right)=n f\left(m \cdot \frac{1}{n}\right)=f\left(\frac{1}{n}+\frac{1}{n}+\cdots+\frac{1}{n}\right)=\left[f\left(\frac{1}{n}\right)\right]^{m}=\left(a^{\frac{1}{n}}\right)^{m}=a^{\frac{m}{n}} .
$$

In the original equation, let $y=0$, we have $f(x) \cdot f(0)=f(x)$, since $f(x)$ is monotonic, we know $f(x)$ is not always zero, thus $f(0)=1=a^{0}$.
In the original equation, let $y=-x=-\frac{m}{n}(n, m \in \mathbf{N})$, then we have $f\left(-\frac{m}{n}\right) \cdot f\left(\frac{m}{n}\right)=f(0)$, i.e., $\quad f\left(-\frac{m}{n}\right)=\frac{1}{f\left(\frac{m}{n}\right)}=\frac{1}{a^{\frac{m}{n}}}$.
Since $f\left(-\frac{m}{n}\right)$ is meaningful, it must be that $a \neq 0$, then it must be that $a>0$.
Thus, we have found the solution of the functional equation in the set of rational numbers as $f(x)=a^{x}(a>0)$.
Since $f(x)$ is monotonic, it cannot always be 1, so $f(x)=a^{x}(a>0, a \neq 1)$ is an exponential function.
When $x=\alpha$ is an irrational number, let $a_{i}, b_{i}$ be the exact values to the $i$-th decimal place of $\alpha$, not approximate values and excess approximate values.
When $f(x)$ is an increasing function, we have $f\left(a_{i}\right)f(\alpha)>f\left(b_{i}\right)$.
And $f\left(a_{i}\right)=a^{a_{i}}, f\left(b_{i}\right)=b^{b_{i}}$, thus we get $f(\alpha)=a^{\alpha}(a>0, a \neq 1)$. Therefore, the solution of the original functional equation is $f(x)=a^{x}(a>0, a \neq 1)$, where $a=f(1)$.