If $b>0$, then the multiples of $b$ when listed from negative to positive, in ascending order, are
$$\cdots,-4 b,-3 b,-2 b,-b, 0, b, 2 b, 3 b, 4 b, \cdots$$

If $b>0$, there exists an integer $q$ such that $q b \leqslant a<(q+1) b$. And when $b<0$, there exists an integer $q$ such that $q b \leqslant a<(q-1) b$. Therefore, we have $a=b q+r$, and $0 \leqslant r<|b|$.

Now we need to prove that there is a unique pair of $q, r$ such that $a=b q+r, 0 \leqslant r<|b|$ holds. Suppose there is another pair $q_{1}, r_{1}$, such that
$$a=b q_{1}+r_{1}, \quad 0 \leqslant r_{1}<|b|$$

holds, then subtracting the two equations, we get
$$0=b\left(q-q_{1}\right)+\left(r-r_{1}\right)$$

which means $-b\left(q-q_{1}\right)=r-r_{1}$. So by definition 1, $b \mid\left(r-r_{1}\right)$. According to lemma 1, $|b||| r-r_{1} \mid$. Since $0 \leqslant r<|b|, 0 \leqslant r_{1}<|b|$, we have
$$\left|r-r_{1}\right|=\left\{\begin{array}{ll} r-r_{1} \leqslant r<|b|, & \text { when } r \geqslant r_{1}, \\ r_{1}-r \leqslant r_{1}<|b|, & \text { when } r<r_{1}. \end{array}\right.$$

From $\left|r-r_{1}\right|<|b|,|b||| r-r_{1} \mid$ and lemma 3, we get $r-r_{1}=0$, which means $r=r_{1}$. By $b \neq 0$ and $b\left(q-q_{1}\right)=r_{1}-r=0$, we get $q-q_{1}=0$, which means $q=q_{1}$.