Reminds of Division Algorithm?

Number Theory Level 3

True or False?

For any integer a a and any positive integer b ≠ 0 b\neq0 , it's always possible to find integers q q and r r such that a = b q + r a=bq+r and ∣ r ∣ ≤ ⌊ b 2 ⌋ |r| \leq \left \lfloor \dfrac{b}2 \right \rfloor .

True False

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Muhammad Rasel Parvej by Muhammad Rasel Parvej , 27, Bangladesh

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1 solution

Recall the Division Algorithm.

For any integer a a and any positive integer b ≠ 0 b\neq0 , it's always possible to find integers Q Q and R R such that a = b Q + R , a=bQ+R, with 0 ≤ R < b 0 \leq R <b .

Now there are two cases.

  • R ≤ 1 2 b R \leq \dfrac{1}{2}b : Then as R ≥ 0 R\geq 0 , ∣ R ∣ ≤ 1 2 b |R| \leq \dfrac{1}{2}b . We have q = Q q=Q and r = R r=R .

  • R > 1 2 b R > \dfrac{1}{2}b : But 0 ≤ R < b 0 \leq R <b , which implies 1 2 b < R < b ⟹ 0 < b − R < 1 2 b ⟹ ∣ b − R ∣ < 1 2 b \dfrac{1}{2}b < R < b \implies 0 < b-R< \dfrac{1}{2}b \implies |b-R|< \dfrac{1}{2}b . Now, a = b Q + R = b Q + b − b + R = b ( Q + 1 ) + ( R − b ) a=bQ+R= bQ+b-b+R=b(Q+1)+ (R-b) . As ∣ b − R ∣ < 1 2 b ⟹ ∣ R − b ∣ < 1 2 b |b-R|< \dfrac{1}{2}b \implies |R-b|< \dfrac{1}{2}b . We have q = Q + 1 q=Q+1 and r = R − b r=R-b .

So the answer is Y e s \boxed{Yes} .

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