Question

Given positive integers a and b. there exist inique integers q and r satisfying a=bq+r, $0\le r<b$.

Answer 1

Let 'a' and 'b' be any positive integers.
Then there exist a unique integers 'q' 'r' such that
$a=bq+r;0\le r\le b$
If $\frac{b}{a}$ then r = 0. Otherwise 'r' satisfies the stronger mearality $0\le r\le b$
Proof: Consider the following arithmetic progression
...a-2b, a-b, a, a+b, a+2b...
Here common difference is 'b' it extends to the both direction.
Let 'r' be the smallest non-negative of this A.P.
Then there exists a non-negative integer 'q' such that
a-bq = r = a = bq+r
As, it is the smallest non-negative integer satisfying the above result.
Therefore, $0\le r\le b$
Thus, we have
$a=bq+r,0\le r\le b$
We shall now prove that $r_1=r$ and $q_1=q$
We have,
$a=bq+r$ and $a=b{q}_1+r_1$
$\Rightarrow bq+r=b{q}_1+r_1$
$\Rightarrow r_1-r=b{q}_1-bq$
$\Rightarrow r_1-r=b(q_1-q)$
$\Rightarrow b{r}_1-r$
$\Rightarrow r_1-r=0$ [Since, 0$\le$ r<b and 0$\le r_1$<b $\Rightarrow 0\le r_1-r\le b]$
$\Rightarrow r_1=r$
$\Rightarrow b{q}_1=bq$
$\Rightarrow q_1=q$
Hence the representation is a = bq+r,0$\le$ r<b