Consider $n\in \mathbb{Z}$ with $n > 0$, we define $n\mathbb{Z}=\{kn \in \mathbb{Z} : k\in \mathbb{Z}\}$. It is easy to see that $n\mathbb{Z}$ is a normal subgroup of the additive group $(\mathbb{Z},+)$. We can then consider the quotient $\mathbb{Z}/n\mathbb{Z}$ which is used in many examples in group theory books.
Now, currently I'm trying to show that for the equivalence relation given by $n\mathbb{Z}$ there are just $n$ distinct cosets, which are exactly:
$$n\mathbb{Z},1+n\mathbb{Z},\dots,(n-2)+n\mathbb{Z},(n-1)+n\mathbb{Z}.$$
To show this, I know that what I must show is that if $x\in \mathbb{Z}$ then there is some $m\in \{0,\dots,n-1\}$ such that $x\sim m$, or in other words, there is some $l\in \mathbb{Z}$ such that $x = m + nl$.
We have then two possibilities: $x$ is a multiple of $n$ or not. If it is a multiple of $n$, then there's $l\in \mathbb{Z}$ such that $x = nl = 0 + nl$. In that case, $x\sim 0$, so that $x\in n\mathbb{Z}$ and we are done.
If $x$ is not a multiple of $n$, then for all $l\in \mathbb{Z}$ we have $x-nl=m(l)$ with $m(l)\neq 0$. Now, in that case $x = m(l)+nl$. So it all boils down to prove that there is $l\in \mathbb{Z}$ such that $m(l)\in \{1,\dots,n-1\}$.
I believe that some argument with the well ordering principle might be the way to go, but I'm unsure on how to prove this. How can we finish this proof following this line of reasoning?
Also, I know that all of this relates to the division algorithm and the idea of quotient and remainder, but I really don't know much about this, and I'm trying to make this proof without any appeal to this. Indeed, after this proof is done I think we can define the $l$ for which $m(l)\in \{1,\dots,n-1\}$ to be the quotient and the corresponding $m(l)$ to be the remainder.
Finding an $l$ such that $1\le m(l)\le n-1$, is not difficult. Let $1\le k \le n-1$,
$$k=k+0\cdot n.$$