Let $a,b$ integers ($b>0$ ). We want to express $a=bq+r$ for some integers $q,r$ with $0\leq r < b$ and that such expression is unique.

Consider the numbers $$\ldots, a-3b, a-2b, a-b,a,a+b,a+2b,a+3b,\ldots$$ From all these numbers, there has to be a smallest non negative one. Let it be $r$ . Since $r=a-qb$ for some $q$ ,¹ we have $a=bq+r$ . And, if $r\geq b$ then $r$ wasn't the smallest non-negative number on the list, since the previous (equal to $r-b$ ) would also be non-negative. Thus $0\leq r<b$ .

So far, we have proved that we can express $a$ as$bq+r$ for some pair of integers $q,r$ such that $0\leq r<b$ . Now we will prove the uniqueness of such expression.

Let $q'$ and $r'$ another pair of integers holding $a=bq'+r'$ and $0\leq r' <b$ . Suppose $r\neq r'$ . Since $r' = a-bq'$ is a number on the list, cannot be smaller or equal than $r$ and thus $r<r'$ . Notice that $$0<r'-r =(a-bq') - (a-bq) = b(q-q')$$ so $b$ divides $r'-r$ which is impossible since $0<r'-r < b$ . We conclude that $r'=r$ . Finally, if $r=r'$ then $a-bq = a-bq'$ and therefore $q=q'$ . This concludes the proof of the uniqueness part.

Footnotes

... ,¹

For example, if $r =a+5b$ then $q=-5$ .