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

step-by-step explanation:

i have no idea

the discriminant is 64 + 4*1*12 = 112

this is prime.

step-by-step explanation: