Example 5 Find the fraction $\frac{a}{b}$ with the smallest denominator such that
$$
\frac{2}{5}<\frac{a}{b}<\frac{7}{9} .
$$
Solution: $\frac{2}{5}<\frac{a}{b}<\frac{7}{9} \Leftrightarrow \frac{9}{7}<\frac{b}{a}<\frac{5}{2}$
$$
\Leftrightarrow \frac{2}{7}<\frac{b-a}{a}<1 \frac{1}{2} \text {. }
$$

By observation, the rational number with the smallest denominator between $\frac{2}{7}$ and $1 \frac{1}{2}$ is clearly 1, so we only need to set
$$
\frac{b-a}{a}=1 \Leftrightarrow\left\{\begin{array} { l } 
{ b - a = 1 , } \\
{ a = 1 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
b=2, \\
a=1,
\end{array}\right.\right.
$$

That is, $\frac{1}{2}$ is the fraction with the smallest denominator we are looking for.
Thus, we have used a so-called clever method and a progressive method to solve a type of competition problem. It is worth noting that problems that can be solved by clever methods are often rare exceptions in a class of mathematical problems.