Step-by-step explanation:
Correct option is
C
0≤r<b
If r must satisfy0≤r<b
Proof,
..,a−3b,a−2b,a−b,a,a+b,a+2b,a+3b,..
clearly it is an arithmetic progression with common difference b and it extends infinitely in both directions.
Let r be the smallest non-negative term of this arithmetic progression.Then,there exists a non-negative integer q such that,
a−bq=r
=>a=bq+r
As,r is the smallest non-negative integer satisfying the result.Therefore, 0≤r≤b
Thus, we have
a=bq1+r1, 0≤r1≤b
