Question Prove that if a and b are positive integers,then there exists a unique integers q and r such that a=bq+r where 0≤r<b

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 Reply

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