Understanding the proof of $a\equiv b\pmod{n}\implies r_n(a)=r_n(b)$

Question

I have seen a proof of the following problem, but I don't fully understand the proof.

Prove that $$a\equiv b\pmod{n}\implies r_n(a)=r_n(b),$$ where $r_n(h)$ means the remainder of $h$ in the division by $n$.

Proof. By the division algorithm, $a=qn+r_a$ and $b=cn+r_b$. Then $$a-b=qn+r_a-cn-r_b=(q-c)n+(r_a-r_b).$$ By hypothesis $a\equiv b\pmod{n}$ i.e. $n\mid a-b$, so it must be $r_a-r_b=0$ i.e. $r_a=r_b$. $\square$

My doubt is in the last part: "so it must be $r_a-r_b=0$".

Question (in red): If $n\mid a-b$ (I understand this) then there exists a $k\in\Bbb{Z}$ such that $a-b=kn$. Hence we would have: $$a-b=\;\;\;\underbrace{kn=(q-c)n}_{\color{red}{\text{How can be equal $k$ and $q-c$?}}}+(r_a-r_b).$$

@RoddyMacPhee I am not seeing why $n\mid a-b\implies r_a-r_b=0$. — manooooh, Dec 14 '19 at 21:32
@Bernard that's what we need to prove... Please look at the comment above (sorry for this misunderstandings). — manooooh, Dec 14 '19 at 21:32

José Carlos Santos · Answer 1 · 2019-12-14T21:39:19.663

2

Note that both $r_a$ and $r_b$ belong to $\{0,1,\ldots,n-1\}$. Therefore,$$r_a-r_b\in\{0,\pm1,\pm2,\ldots,\pm(n-1)\}.$$But the only element of this set which is a multiple of $n$ is $0$.

edited Dec 14 '19 at 21:39

answered Dec 14 '19 at 21:30

José Carlos Santos

427,504

Please strive not to add more dupe answers to dupe FAQs – Bill Dubuque Dec 14 '19 at 21:43

score 1 · Answer 2 · answered Dec 14 '19 at 21:30

1

Hint: Suppose $k$ is not $q-c$ then $(k-(q-c))n=r_a-r_b$ hence $n$ divides $r_a-r_b$ but what does the division algorithm tells you about them?

answered Dec 14 '19 at 21:30

Phicar

14,722

This is a great help! I don't know what the division algorithm tells me. If $k\neq q-c$ then $n\mid r_a-r_b$, then..? – manooooh Dec 14 '19 at 21:34
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the algorithm tells you that $0\leq r_a,r_b<n$. – Phicar Dec 14 '19 at 21:35
Oh yes, you are right. Then we have $0\leq r_a<n$ and $0\leq r_b<n$. What is the next step? – manooooh Dec 14 '19 at 21:36
Then we have $0-0\leq r_a-r_b<n-n=0\implies r_a-r_b=0$ right? – manooooh Dec 14 '19 at 21:37
Nop, you can not subtract like that, but then how can you get a number greater or equal than $n$ by subtracting them?? – Phicar Dec 14 '19 at 21:38
Why do we need to find a number greater or equal than $n$? – manooooh Dec 14 '19 at 21:41

Understanding the proof of $a\equiv b\pmod{n}\implies r_n(a)=r_n(b)$

2 Answers2