Let $\sim_n$ be the usual equivalence relation of congruence modulo $n$ in $\mathbb{Z}$, i.e., for $a,b\in\mathbb{Z}$, $a\sim_nb\Leftrightarrow a-b=k\cdot n$ for some $k\in\mathbb{Z}$.
For $n=0$ the element in the equivalence class of, say, $a\in\mathbb{Z}$ is just $a$ itself. So to every element of $\mathbb{Z}/\sim_0$ corresponds one, and only one, element of $\mathbb{Z}$.
My question is:
What is mathematically more correct, to say that the two sets $\mathbb{Z}/\sim_0$ and $\mathbb{Z}$ are equal or to say that those rings are isomorphic?
We should state the question more general:
Given a commutative ring $R$ with identity and an ideal $I \subseteq R$, we can always build the quotient $R/I$. Now we can ask: Is $R/\{0\}$ equal to $R$ or are they just isomorphic?
Point of view 1: Let us define $R/I = \{a + I \mid a \in R\}$ together with the usual addition and multiplication of cosets. Then we have $R/\{0\} = \{a + \{0\} \mid a \in R\} = \{\{a\} \mid a \in R\}$. This is not equal to $R$ as $\{a\} \neq a$, but the ring homomorphism $R \to R/\{0\}, a \mapsto \{a\}$ is always an isomorphism.
Point of view 2: We define $R/I$ a bit more categorical as follows: For $R/I$ we choose some ring together with a surjective ring homomorphism $\pi: R \to R/I$ such that $\text{ker}(\pi) = I$. The process of choosing could go like that: If $I \neq \{0\}$ we let $R/I = \{a + I \mid a \in R\}$ and $\pi: R \to R/I$ be the natural projection. If $I = \{0\}$ we let $R/I = R$ and $\pi: R \to R/I$ the identity. Now the identity $\text{id}: R \to R$ is a surjective ring homomorphism with kernel $\{0\}$, so this is indeed well defined.
I prefer point of view $2$: In fact, just using this definition (and not the construction!) allows us to prove every statements about quotient rings we want to have. Also it's easier to use, as we don't get confused by cosets, if we don't want to. The first definition just exists to prove that a ring with the properties of the second definition does always exist.