This is exercise 9.2 from textbook Analysis I by Amann/Escher.
For $n \in \mathbb{Z}$, $n \mathbb{Z}$ is an ideal of $\mathbb{Z}$, and so the quotient ring $\mathbb{Z}_{n} :=\mathbb{Z} / n \mathbb{Z}$, $\mathbb{Z}$ modulo $n$ is well defined. Show that, for $n \in \mathbb{N}^{ \times}$, $\mathbb{Z}_{n}$ has exactly $n$ elements. What is $\mathbb{Z}_{0}$?
Here $\mathbb{N}^{\times} := \mathbb{N} - \{0\}$.
I would like to confirm if my attempt is fine or contains logical flaws, and if it is formal enough to constitute a proof. Thank you for your help!
My attempt:
Lemma: For all $p \in \mathbb{Z}$, $n \in \mathbb{N}^{\times}$, there exist unique $m$ and $0 \le r < n$ such that $p = mn +r$. We call $r$ the remainder when $n$ divides $p$.
Proof: Assume there exist $m_1, m_2$ and $0 \le r_1, r_2 < n$ such that $p = m_1 n +r_1 = m_2 n +r_2$. Then $(m_1 - m_2)n = r_2 - r_1$. If $m_1 > m_2$ then $r_2 = (m_1 - m_2)n + r_1 > n$, which is a contradiction. Similarly, the case $m_1 < m_2$ is impossible. As such, $m_1 = m_2$ and thus $r_1 = r_2$. The uniqueness then follows.
We prove the existence by induction in case $p \in \mathbb{N}$. Since $0 = 0n +0$, the lemma holds for $p = 0$. Let it hold for $p$. Then, for all $n \in \mathbb{N}^{\times}$, there exist $m$ and $0 \le r < n$ such that $p = mn +r$. We have $p + 1 = mn +r + 1$. If $r = n-1$ then $p + 1 = mn +(n -1) +1 = (m+1)n + 0$. If $r < n-1$, then $p+1 = mn + (r+1)$ where $r+1 < n$. As a result, the lemma holds for $p+1$. By recursion theorem, the lemma holds for all $p \in \mathbb{N}$. For $p < 0$, let $p' := -p > 0$. Then, for all $n \in \mathbb{N}^{\times}$, there exist $m$ and $0 \le r < n$ such that $p' = mn +r$. If $r=0$ then $p=(-m)n+0$. If $r>0$ then $p = -mn -r = (-m - 1)n +(n-r)$ where $0 \le n - r < n$. The existence then follows.
For $p,q \in \mathbb{Z}$, $p \sim q$ if and only if $q = p + nt$ for some $t \in \mathbb{Z}$. By lemma, there exist unique $m$ and $0 \le r < n$ such that $p = mn +r$. It follows that $q = (mn+r)+nt = (m+t)n + r$. Then $p \sim q$ if and only if $p$ and $q$ have the same remainder when they are divided by $n$. As such, each equivalence class in $\mathbb{Z}_n$ is characterized by the remainder when $n$ divides its elements.
By lemma, when $n$ divides $p \in \mathbb{Z}$, the remainder belongs to $\{0,\ldots,n-1\}$. As a result, $|\mathbb{Z}_n| = |\{0,\ldots,n-1\}| = n$.
Clearly, $\mathbb{Z}_0 = \mathbb{Z}$.
