Why is $\mathbb{Z}_n$ defined to contain equivalence classes instead of integers?

Question

I.e., $\mathbb{Z}_4=\{[0],[1],[2],[3]\}$ instead of $\{0,1,2,3\} $.

We just went over this in my discrete math class, so I'm wondering what higher-level reasons mathematicians had to choose the former instead of the latter.

Answer 1

As was said in the comments, how the elements of your system of numbers looks does not really matter, as long as the operations behave in the same way, so really both options are legitimate.

It turns out that defining $\mathbb Z_n$ as equivalence classes makes the operations behave a little bit nicer. Again, if we restrict our view to just $\mathbb Z_n$ this does not change anything at all, but if we see this set as somehow "built out of $\mathbb Z$, then the definitions seem a bit more 'natural'. In particular, it might be less tedious to prove that properties like associativity hold.

In particular, if $[n]$ and $[m]$ are equivalence classes, then the sum $[n + m]$ can be defined in an intrinsic way to be the set of all $a + b$ for $a \in [n]$ and $b \in [m]$. For example, the sum $[1] + [2]$ in $\mathbb Z_4$ will contain $1 + 2 = 3$, as well as $1 + 6 = 5 + 2 = 7$ or $-7 + 6 = -1$. You can prove that this set of sums for any two equivalence classes is just the equivalence class of a sum of a pair of representatives of these equivalence classes. In this way, the definition of addition is more "natural" or "intrinsic" and does not depend on any complex properties of $\mathbb Z$. (It does depend on addition, but not on the ability to take numbers modulo each other)

You should note that the story is not as simple in the case of multiplication: We will still have that $[a]\cdot [b]$ (the set of all products of an element in $[a]$ with an element in $[b]$) is contained in $[a\cdot b]$, but the other direction need not necessarily be true. Still, with a bit more work, you can find an equally natural description.

One final reason why you would want to choosing your elements of $\mathbb Z_n$ to be specific numbers in $\mathbb Z$ is that the precise choice of these elements would be somewhat arbitrary (You chose the numbers $0, \ldots, n-1$, but we could equally well have chosen the numbers $-\frac n2+1, \ldots, 0, \ldots, \frac n2$ if $n$ is even, and something similar if $n$ is odd). Choosing equivalence classes as your elements is somehow the "canonical" way to do it.

Answer 2

If the elements are integers then you might be tempted to think that arithmetic there is integer arithmetic - but of course it's not. It's modular arithmetic.

That said, we often use the integers from $0$ to $n-1$ as default representations for their cosets and don't bother with the square brackets. The context tells the reader how the arithmetic works.

Finally, there are some classic proofs in elementary number theory that work because you look at particular representatives of the congruence classes - for example, Gauss's lemma in the proof of quadratic reciprocity.

Answer 3

The set of equivalence classes is a partition of the set of integers. In other words, the set of integers $\mathbb{Z}$ can be partitioned into sets $A_1,\ldots,A_n$ such that

1. $\bigcup_{i=1}^n A_i = \mathbb{Z}$,
2. $A_i\cap A_j = \emptyset \:\forall i,j: i\neq j$
3. $A_i\neq \emptyset$ for every $i$

Therefore, the set of equivalence classes must represent every integer. This is why the act of listing arbitrary integers is not sufficient to describe the structure of the equivalence class (even though, through abuse of notation, we often see integers used in place of equivalence classes in textbooks)