I'm trying to prove that $\mathbb{Z}/n\mathbb{Z} = \{[0], [1], \ldots, [n-1]\}$, where $n \geq 1$ is fixed. Here is my attempt.
First, each of $[0], [1], \ldots, [n-1]$ is an equivalence class. If $x \in [r]$ for $0 \leq r \leq n-1$, then the remainder when $x$ is divided by $n$ is $r$, i.e., $x = nq + r$. But then $x - r = nq$, so $x \equiv r \text{ (mod $n$)}$. By the division algorithm, given $m \in \mathbb{Z}$, there exist unique $q,r$ such that $m = qn + r$ where $0 \leq r \leq n-1$. Then $m - r = qn$, so $m \in [r]$, so these equivalence classes are exhaustive. Finally, we show that each of these equivalence classes are distinct. Given $0 \leq i < j \leq n-1$, suppose that $[i] = [j]$. Then there exists $x \in [i] = [j]$, so $x = mn + i$, $x = pn + j$ for $m,n \in \mathbb{Z}$, so $mn + i = pn + j$, so $mn - pn = i - j$, so $(m-p)n = i - j$, so $i \equiv j \text{ (mod $n$)}$, a contradiction.
How does this proof look? Have I made any errors or omitted any steps? It seems to me that demonstrating that these equivalence classes are exhaustive establishs the left inclusion, $\mathbb{Z}/n\mathbb{Z}$. I don't know exactly what the right inclusion would look like here, but the right set could definitely be truncated if $[i] = [j]$ for $i \neq j$, so I need to check that the classes are distinct.
I would appreciate any feedback, including a better way to think of how to prove (or how to organize a proof) of something like this.
Per Bill Dubuque's comment below, I am updating my answer to be as general as possible. By definition, $\mathbb Z / n \mathbb Z$ is the set of equivalence classes of the integers $\mathbb Z$ modulo the equivalence relation $a \sim b$ if and only if $a - b$ is divisible by $n.$ Consequently, the equivalence classes $a + n \Bbb Z$ and $b + n \Bbb Z$ are equal if and only if $a - b = qn$ for some integer $q.$ By the Division Algorithm, we may write $a = qn + r$ for some integers $q$ and $0 \leq r \leq n - 1.$ Observe that $a + n \Bbb Z = r + n \Bbb Z.$ Ultimately, we find that $$\Bbb Z / n \Bbb Z = \{0 + n \Bbb Z, 1 + n \Bbb Z, \dots, (n - 1) + n \Bbb Z\}$$ because the only possible remainders modulo $n$ are $0, 1, \dots, n - 1.$
Originally, this post included the following additional comments.
"Overall, it seems to me that there is some confusion about the 'meaning' of $\Bbb Z / n \Bbb Z$: you should explain in what context you are viewing $\Bbb Z / n \Bbb Z.$ (Is it a set, a group, or a ring?) On the matter of your argument, it would make more sense to me if you defined the equivalence relation $\sim$ and subsequently stated what $[r]$ means in terms of $\sim.$"