proof of uniqueness regarding division with remainder

Question

Let $m,n \in N$ with n>0, then there exists $q,r \in N$ such that $m=qn+r$ with $0\le r < n$

Show for any given $m,n$ with $n>0$ the $q$ and $r$ are unique. So if $m=q_1n+r_1=q_2n+r_2$ with $0\le r_1\lt n$ and $0\le r_2< n$ then $q_1=q_2$ and $r_1=r_2$.

I tried contradiction, so I assumed $m=q_1n+r_1=q_2n+r_2$ with $0\le r_1< n$ and $0\le r_2< n$ and $q_1\ne q_2$ or $r_1\ne r_2$.

Then I tried to show a contradiction in the three cases which make $q_1\ne q_2$ or $r_1\ne r_2$ true, is this the right thing to do here?

so with one of them not equalland one of them equal I just used $m=q_1n+r_1=q_2n+r_2$ and its quite easy to show a contradiction.

with when they are both not equal

$-n<r_2-r_1<n$

implies $-1<q_1-q_2<1$

which contradicts the fact that $q_1\ne q_2$ and $q_1,q_2\in Z$

So is this a valid proof? is there a quicker way to prove this? Thanks

Answer 1

You are doing too much.

The case of $r_1\ne r_2$ Is sufficient for the contradiction.

Since if $r_1=r_2$ then $q_1=q_2$ follows.

Otherwise your proof is correct. You have the inequalities of the remainder backward so fix it first.

Answer 2

You don't really need to proceed by contradiction:

Suppose $m=q_1n+r_1=q_2n+r_2$, with $0\le r_1,r_2<n$. Suppose for instance that $r_1\le r_2$. Then we deduce that $$0\le r_2-r_1=(q_1-q_2)n,$$ in other words, $r_2-r_1$ is a multiple of $n$. However we have $$0\le r_2-r_1\le r_2<n.$$ The only multiple of $n$ which is both $\ge 0$ and $<n$ is $0$. Hence $r_2=r_1$. It is now easy to obtain $q_1=q_2$.