With modular arithmetic, it seems possible to define the addition operation without imposing equivalence classes. That is, with the set of remainders when dividing by $n$ as $\{0, …, n – 1\}$, we can define the addition between members of this set to be the usual integer sum in $\mathbb{Z}$, reduced modulo $n$. My university notes say that, with this definition, addition is well-defined but that its properties are difficult to glean. As a result, equivalence classes are imposed and addition occurs between these classes. My questions are: (1) why is it difficult to glean the properties of the original addition (please provide examples to make it clearer) and (2) how does the introduction of equivalence classes solve this problem? Below is an excerpt from the notes I am referring to:
Another group with a different flavor is $(\mathbb{Z}/(n\mathbb{Z}), +)$, the integers modulo $n ≥ 2$ integer: as a set, this can be identified with $\{0, 1, . . . , n − 1\}$ (the remainders when dividing by $n$ in integers), and addition gives the usual sum in $\mathbb{Z}$, reduced modulo $n$, so e.g. in $(\mathbb{Z}/(5\mathbb{Z}), +)$, $2 + 4 = 1$. It is less confusing though to write $\{[0], . . . , [n − 1]\}$ for the set, and $[2] + [4] = [1]$ then; below we see why this is a good notation.
With this definition, addition is well defined on $(\mathbb{Z}/(n\mathbb{Z}), +)$, but it can be quite awkward to check its properties. It is usually better to proceed as follows: we partition (divide into disjoint sets) $\mathbb{Z}$ into equivalence classes, and say that $a$ and $b$ are in the same class, or $a ∼ b$, if $a − b$ is divisible by $n$, i.e. if $a$ and $b$ differ by a multiple of $n$. Then there are $n$ equivalence classes, namely exactly the remainders when dividing by $n$. We write the equivalence class of $a ∈ \mathbb{Z}$ as $[a]$…
Best Answer
Consider, for example, the associative property. Without using equivalence classes (at least implicitly*), to check that $(a+b)+c = a + (b + c)$ one has to consider the results of four different additions. If the definition of addition in modular arithmetic is given by the remainder of the actual sum on division by $n$, then it can either be the sum itself, or the sum minus $n$. Thus there are two cases to check for each addition. Since there are four additions, that amounts to $2^4=16$ different cases to check. Some of these will be quite similar to each other, but it is still rather a cumbersome process to wade through all $16$ possibilities.
On the other hand, if we are talking about equivalence classes, after showing that $[a]+[b]:=[a+b]$ is well-defined, we get associativity almost for free, since \begin{align*} ([a]+[b])+[c] &= [a+b]+[c] = [(a+b)+c] = [a + (b+c)] \\ &= [a] +[b+c] = [a]+ ([b]+[c])\text{,} \end{align*} with every equation following from the definition of modular addition except the middle one, which follows from associativity of regular addition.
*Update: In the comments, user21820 correctly points out that the equivalence classes do not need to be explicitly defined in order to establish associativity as in the argument above - one could indeed simply define $[a]$ to be $a$ reduced mod $n$, and then prove that $[a+b]=[[a]+[b]]$.
Here equivalence classes are still at play implicitly, as any map $f\colon X\rightarrow Y$ induces an equivalence relation on its domain with equivalence classes of the form $f^{-1}(\{y\})$ under that map, and in this case, the map $a\mapsto a \mod n$ induces the very equivalence relation on $\mathbb Z$ that we use to define modular arithmetic.