Proof: If $\frac{a}{b}$ can be expressed as a pure repeating decimal, then by $0<\frac{a}{b}<1$ and Definition 2, we have
$$\frac{a}{b}=0 . a_{1} \cdots a_{t} a_{1} \cdots a_{t} a_{1} \cdots a_{t} \cdots$$

where $a_{1}, \cdots, a_{t}$ are non-negative integers no greater than 9, but at least one $a_{i} \geqslant 1$ in $a_{1}, \cdots, a_{t}$. Therefore,
$$\frac{10^{i} a}{b}=10^{t-1} a_{1}+\cdots+a_{t}+0 . a_{1} \cdots a_{t} a_{1} \cdots a_{t} a_{1} \cdots a_{t} \cdots$$

Subtracting (2) from (3) yields
$\frac{10^{t} a}{b}-\frac{a}{b}=10^{t-1} a_{1}+\cdots+a_{i}$, hence we get
$$a\left(10^{t}-1\right)=b\left(10^{t-1} a_{1}+\cdots+a_{t}\right)$$

Since $10^{t-1} a_{1}+\cdots+a_{i}$ is a positive integer, $(a, b)=1$ and (4), we have
$$10^{t}-1=b m$$

where $m$ is an integer. By $(b, 1)=1$ and (5), we have $\left(b, 10^{t}\right)=$ 1, thus $(b, 10)=1$.