In Gallian's Contemporary Abstract Algebra, he defines $\mathbb{Z}_n$ as the set (see 10e, Ch.2: p. 39 – 40)$$\mathbb{Z}_n := \{0, 1, 2, 3, \dots, n-1 \}.$$
To be clear, at this point in the text, equivalence classes modulo $n$ have been introduced, and Gallian takes care to denote them by $[a]$. The reader would have no reason to believe that the 0, 1, 2, 3, etc. in the set above are anything but elements of the integers (I've searched the book for comments concerning this issue).
In Hungerford's Introduction to Abstract Algebra, $\mathbb{Z}_n$ is defined using equivalence classes modulo $n$: $$\mathbb{Z}_n := \{ [0], [1], [2], \dots, [n-1] \} \not \subset \mathbb{Z} \quad \text{where} \quad [a] = \{ b \in \mathbb{Z} \, \vert \, b = a \bmod n\}.$$
In both texts, the authors go on to say that $\mathbb{Z}_n$ forms a group under addition modulo n, though I suppose in the first, the "addition" has mod $n$ built in, while in the second, this gets buried in the definition of addition for equivalence classes mod $n$; that is
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For the first: $+: \mathbb{Z}_n \times \mathbb{Z}_n \to \mathbb{Z}_n$ is defined by $a+b = (a+b) \bmod n$
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For the second: $+: \mathbb{Z}_n \times \mathbb{Z}_n \to \mathbb{Z}_n$ is defined by $[a] + [b] = [a +_\mathbb{Z} b]$
Question: Is it important to distinguish between these definitions, or is it just a matter of perspective? What benefits are there in defining $\mathbb{Z}_n$ using equivalence classes rather than as just as a subset of the integers (pedagogically or otherwise)?
Best Answer
The benefit of the equivalence class point of view is its vast generality and its abstract simplicity (one person's "this gets buried in other stuff" is another person's "abstract simplicity").
Since you have an abstract-algebra tag, perhaps you have learned some group theory, including normal subgroups and quotient groups.
In your post, think of $G=\mathbb Z$ as a group under the operation of addition and $n\mathbb Z = \{nk \mid k \in \mathbb Z\}$ as a normal subgroup. The "equivalence classes" are defined by the relation $a \sim b$ if and only if $b-a \in n\mathbb Z$.
This definition generalizes to any group and normal subgroup.
That is, let $G$ be a group, with group operation denoted $g \cdot h$ and inverse operation denoted $g^{-1}$ (notice, I have switched to multiplicative notation, but this is just a notational change). Let $N < G$ be a normal subgroup. We can define an equivalence relation: $g \sim h$ if and only if $g^{-1} h \in N$. Letting $[g]$ be the equivalence class of $g \in G$, we can then define a group operation $[g] \cdot [h] = [g \cdot h]$. There's plenty of details to check: that the relation $g \sim h$ does, indeed, satisfy the axioms of equivalence relations (true whether or not the subgroup $N$ is normal); next that the operation $[g] \cdot [h] = [g \cdot h]$ is well-defined (this is where normality of $N$ is needed); and then one must work to verify the group axioms for this new operation.
The resulting group is called the "quotient of $G$ modulo $N$" and is sometimes denoted $G/N$. So the example of a quotient in your post is often denoted more formally as $\mathbb Z / n \mathbb Z$.
The outcome is that one obtains a very general and spectacularly powerful tool in group theory.
Let me also comment on your question of whether it is important to distinguish between the two definitions.
I would say "not really".
Given any equivalence relation $\sim$ on any set $X$, one common operation is to "choose representatives", i.e. choose one element in each equivalence class. This gives a subset of $X$ that is in one-to-one correspondence with the set of equivalence classes. In the case of a group and a normal subgroup, once a set of representatives has been chosen, one can then use transport of structure to put a unique group operation on the chosen set of representatives that makes the one-to-one correspondence into a group isomorphism. That's exactly what's going on in your question.
Sometimes there seems to be an "obvious" or a "natural" choice of representatives, and for your example the subset $\{0,1,2,...,n-1\}$ seems like the most obvious choice, although you will see that sometimes people choose $\{1,2,...,n-1,n\}$. For $\mathbb Z_3$ I think that $\{-1,0,+1\}$ is also a nice choice.