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I'm learning Number Theory on my own for competitive maths. The textbook I'm using is the part of the Olympiad syllabus in my country, so the problems are picked and adapted from other textbooks. I'm not quite sure where this particular one came from.

The book presents the proof for the division algorithm and I am asked to prove this case.

Show that if $a, b \in Z $ and $a \gt 0$ then there exists a unique pair of $q, r \in Z$ that satisfies $b = aq + r$ and $2a \le r \lt 3a$

Thanks for the help!

Shawanwit Poomsa-ad
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    Welcome to Math SE. If $a = 1$, there's no integer $r$ where $2 \lt r \lt 3$. Also, if $a \mid b$, then $a \mid r$, so there's no $2a \lt r \lt 3a$. I suspect it should be $2a \le r \lt 3a$ instead. – John Omielan Feb 20 '21 at 04:51
  • @JohnOmielan Thank you John Omielan, I have spotted typos in this textbook before. Could you please provide the proof for the case that you mentioned? – Shawanwit Poomsa-ad Feb 20 '21 at 08:11
  • You're welcome. For me, or anyone else here, to best help you, please first edit into the question more context, in particular what you've tried using my suggested correction, and anything you had difficulty with. – John Omielan Feb 20 '21 at 08:17
  • @JohnOmielan Ok I'll get to that. – Shawanwit Poomsa-ad Feb 20 '21 at 08:32
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    As I wrote in this dupe, if you already know the Division Algorithm then simply apply that to write $,a−2b=qb+r$ with $,0\le r<b,,$ then add $,2b,$ to both sides. More generally we can choose any consecutive sequence of $b$ integers as a complete set of reps for remainders $!\bmod b,$ (provable via the same sort of shift conjugation to standard reps) – Bill Dubuque Feb 20 '21 at 08:53
  • Special case of the result on arithmetic progressions being a complete residue system in this answer in the linked dupe. Ask questions here if that is not clear. – Bill Dubuque Feb 20 '21 at 08:58
  • There is a typo: one of your inequalities on $,r,$ should not be strict (probsbly $,2a\le r < 3a,$ is intended). Also note $,b,a,$ are reversed in the dupe (and the the above excerpted comment). – Bill Dubuque Feb 20 '21 at 09:11
  • @BillDubuque Thank you, I'll look into the post. – Shawanwit Poomsa-ad Feb 21 '21 at 10:05

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