Prove that $r_1 = r_2$ iff $n | (b - a)$ I need to know if I'm clear in my proof since I will have to present the answer to my class.
Here's the full question: Let $n$ be a fixed positive integer. Then for any integers $a$ and $b$, let $r_1$ be the remainder when $a$ is divided by $n$ and let $r_2$ be the remainder when $b$ is divided by $n$.  Then $r_1 = r_2$ if and only if $n|(a-b)$
Proof:
(=>) Suppose $a \equiv b (\mod n)$, then $a = nq_1 + r_1$ for $0 < r_1 < n$ and $b = nq_2 + r_2$ for $0 < r_2 < n$.
Now, $a - b = (nq_1 + r_1) - (nq_2 + r_2) = n(q_1 - q_2) + (r_1 - r_2)$
So, $n|(a-b)$.
(<=) Suppose $n|(a-b)$, then there exists $q_1, q_2, r_1, r_2 \in \mathbb{Z}$ such that $a = nq_1 + r_1$ and $b = nq_2 + r_2$.
Consider $(a-b)$ as stated above.
Then $-n < -r_2 \le 0$ and $0 \le r_1 < n$. $0-n < r_1 - r_2 < n , -n < r_1-r_2 < n$ and since $r_1-r_2$ is a multiple of $n$ then $r_1-r_2 = 0$ and Hence $r_1 = r_2$.
I just need to make sure I'm clear on my explanation.  I'm a little unsure of my converse statement because I feel like I need to say more.
 A: A few observations,


*

*For the forward direction you never relate $r_1$ and $r_2$. As far as your audience is concerned these numbers could be different which would mean that you haven't yet established that $n\mid (a-b)$. So you need to make more explicit what $a\equiv b (\mod n)$ tells us about the relationship between $r_1$ and $r_2$. 

*The phrasing in your writing makes it seem as though $a \equiv b (\mod n)$ implies that $a=n q_1 + r_1$ and $b=nq_2+r_2$. There is a similar phrasing in that suggests that $n\mid (a-b)$ implies the existence of $q_1,q_2,r_1,r_2$. These exist because of the division algorithm which holds for all integers and has nothing to do with thre relationship between $a$ and $b$. Rather than saying "then there exists" (which suggests what you are about to say is an implication of the previous statement) you should say "The division algorithm tells us that..."

*In the second part of the proof make sure to state how you know that $r_1-r_2$ is a multiple of $n$. The audience shouldn't have to take any of this on faith.
A: If $a=d_1n+r_1$ and $b=d_2n+r_2$ and $0\leq r_1,r_2<n$ then 
$$a \bmod n= r_1, b \bmod n=r_2\\a\equiv b \pmod n\text{ iff } r_1=r_2$$
Shouldn't be very complicated
