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Prove that if $a$ and $b$ are integers, with $b > 0$, then there exist unique integers $q$ and $r$ satisfying $a = qb + r$, where $2b \le r < 3b$.

In this post, I am only concerned with the part of proving existence and not uniqueness of $q$ and $r$. Here is my proof of existence of $r$:

Let $S = \{a - xb \mid x \in \mathbb Z~ , ~2b \le a - bx < 3b \}$. If possible, let $S$ be empty. So, $$\forall x \in \mathbb Z ~,~ (a - xb < 2b ~\vee ~ a - xb \ge 3b)$$ i.e. $$\forall x \in \mathbb Z ~,~ \left(x > \frac a b - 2 ~\vee~ x \le \frac a b - 3 \right) \tag{1}$$

However, in the interval $\left(\dfrac a b - 3~ , ~ \dfrac a b - 2\right]$ there exists an integer such that $(1)$ becomes false. Since we have reached a contradiction, therefore our assumption of $S$ being empty is false and so it is non- empty which proves the existence of $q$ and $r$.

Is my proof legit?

Parth
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    if you already know the Division Algorithm then simply apply that to write $,a-2b = qb + r,$ for $,0\le r < b$ 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 Sep 18 '17 at 19:54
  • @Bill Dubuque Yes I do know the division algorithm. Its just that I have seen a similar proof of division algorithm and wanted to make a proof along the same lines. – Parth Sep 19 '17 at 04:02
  • Would the downvoter mind explaining the reason for downvoting? Maybe I can improve the post. – Parth Sep 19 '17 at 04:03
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    You would also need to prove that every interval $(r,r+1]$ contains an integer. Usually in number theory the division algorithm is proved before a (rigorous) development of fractions, and the proof is from first principles using induction, e.g. by showing that $,a-b\Bbb Z,\cap, \Bbb N,$ is nonempty so has a least element $,r,$ (necessarily $,r<b,$ else $r-b$ would be a smaller such natural). The same method works in your case - as I said above. I have no idea why your question was downvoted. – Bill Dubuque Sep 19 '17 at 14:29
  • The standard proof (e.g. here in the dupe) still works if we replace the standard remainder interval $[0,b!-!1]$ by any sequence of $b$ consecutive integers (here $[2b,3b!-!1]$). Equivalently, shift the standard remainders by $2b$ by subtracting $2$ from the quotient: $,qb+r = (q!-!2)b+ 2b+r.\ \ $ – Bill Dubuque Nov 14 '24 at 19:53

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