Question

proof of Euclid division lemma.​

Answer 1

Proof of Euclid division lemma:

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.

Answer 2

Consider the following arithmetic progressiona – 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.