Question

Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 <_r>b

Answer 1

Euclids division lemma : Let ‘a’ and ‘b’ be any two positive integers. Then there exist unique integers ‘q’ and ‘r’ such that

a = bq + r, 0 ≤ r ≤ b.

If b | a, then r=0. Otherwise, ‘r’ satisfies the stronger inequality 0 ≤ r ≤ b.

Proof : Consider the following arithmetic progression

…, 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 the 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 above result. Therefore, 0 ≤ r ≤ b

Thus, we have

a = bq1 + r1 , 0 ≤ r1 ≤ b

We shall now prove that r1 = r and q1 = q

We have,

a = bq + r and a = bq1 + r1

⇒ bq + r = bq1 + r1

⇒ r1 – r = bq1 – bq

⇒ r1 – r = b(q1 – q)

⇒ b | r1 – r

⇒ r1 – r = 0 [ since 0 ≤ r ≤ b and 0 ≤ r1 ≤ b ⇒ 0 ≤ r1 - r ≤ b ]

⇒ r1 = r

Now, r1 = r

⇒ -r1 = r

⇒ a – r1 = a – r

⇒ bq1 = bq

⇒ q1 = q Hence, the representation a = bq + r, 0≤ r ≤ b is unique.