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​

Answers

  1. 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

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