Let $d=(a, b)$, in the Euclidean algorithm of Theorem 3 from the previous section, take $u_{0}=a, u_{1}=b$ (here we assume $b \neq 0$), then by the properties of divisibility, it can be known that $d\left|u_{2}, d\right| u_{3}, \cdots$, $d \mid u_{k+1}$. Therefore, $d \leqslant u_{k+1}$.

Conversely, by the properties of divisibility, it can be known that $u_{k+1}\left|u_{k}, u_{k+1}\right| u_{k-1}, \cdots, u_{k+1} \mid u_{1}$, $u_{k+1} \mid u_{0}$, i.e., $u_{k+1}$ is a common divisor of $a$ and $b$. Therefore, $u_{k+1} \leqslant d$.

The above discussion shows: $d=u_{k+1}$. Now, by reversing the equations in the Euclidean algorithm, we can see that
$$u_{k+1}=u_{k-1}-u_{k} q_{k-1}=u_{k-1}-\left(u_{k-2}-u_{k-1} q_{k-2}\right) q_{k-1}=\cdots$$

We successively express $u_{k+1}$ as a linear combination of $u_{k-1}$ and $u_{k}$; express $u_{k+1}$ as a linear combination of $u_{k-2}$ and $u_{k-1}$; and so on, until we express $u_{k+1}$ as a linear combination of $u_{0}$ and $u_{1}$. Therefore, there exist integers $x, y$ such that (1) holds.

Similarly, for more integers $a_{1}, a_{2}, \cdots, a_{k}$, the same conclusion holds.

If $(a, b)=1$, then $a$ and $b$ are said to be coprime. According to the above theorem and the properties of divisibility, it can be known that
$(a, b)=1 \Leftrightarrow$ there exist $x, y \in \mathbf{Z}$ such that $a x+b y=1$.