**Proof:** By division algorithm, there exist unique integers q′ and r′, such that

a = q′ b + r′, 0 ≤ r′ < b

⟹ a = q′ b + r′ + 2b – 2b

⟹ a = q′ b – 2b + r′ + 2b

⟹ a = (q′ – 2)b + r′ + 2b

Let, q = q′ – 2, r = r′ + 2b

*∴ r, q unique integers.*

Since 0 ≤ r′ < b, then

2b ≤ r′ + 2b < b + 2b

⟹ 2b ≤ r < 3b ** (proved)**