Question

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