It's rather intuitive that $[x]_m \oplus [y]_m = [x+y]_m $ and $[x]_m \otimes [y]_m = [x \cdot y]_m $. I was wondering if someone could give me a formal explanation of why $\oplus$ and $\otimes$ 'inherit' the properties of $+$ and $\cdot$. In other words, why is it possible to go from $[\cdots]_m \oplus [\dots]_m $ to $[\cdots+\dots]_m $.
Thanks in advance!
On
The computations: $(a+km) + (b+k'm)= a+b + (k+k')m$, for any $a,b,k,k'\in \mathbb{Z}$,
mean that you can define an operation $\oplus$ on $\mathbb{Z}/m\mathbb{Z}$ such that $$\forall a,b\in\mathbb{Z},\ [a]_m\oplus [b]_m =[a+b]_m$$
On
The key is that you define the operations in this way. Those equivalence classes do not automatically come with any operations on them, but if you define them as corresponding operations on elements, it kind of makes sense. And the formal definition for this kind of sense would be the ring axioms: you'd formally prove that the operations defined in this way do satisfy all the ring axioms, so the data structure with these operations in place forms a ring. You'd also prove that the operations are well defined, in the sense that it doesn't matter which representatives you take, you still get the same equivalence class as a result.
You might wonder whether there is any other way to define these operations to obtain a ring without just inheriting the properties from the natural numbers. For any specific $m$ this is certainly the case, since you could just "relabel" all your ring elements. So the fact that inheriting operations from the natural numbers feels kind of natural is a combination of it having the desired properties, and it being simple to express. The latter is not a strict formal concept, so you'd be hard pressed to show that the operations defined in this way are the reasonable way to do things, in less you formalize further assumptions of what you'd consider reasonable.
I'm going to assume that you mean that this is a "natural" definition for addition and multiplication of reside classes.
By this, I'm assuming you mean that despite the "naturalness" of the definition, you still have some doubts about it.
What kinds of doubt can you have about a definition? if I define a "special prime" to be a negative number that's prime, you might doubt that because
You don't see any value in the definition -- you've never encountered a situation where you needed a word for this thing, or
You notice that the definition is vacuous: the set of special primes is empty.
(Arguably these two are related.)
But it's possible to doubt a definition for another reason: it might not be logically coherent. For instance, if I define a "prent" as "the unique person from whom you inherit genetic material", you might complain that because you have two parents, there's no such thing as a prent. And I might note that you also have 4 grandparents, from whom you inherited genetic material, albeit indirectly. In this case, my definition of "prent" is just nonsense.
There's a particular form of nonsense that's especially pernicious, which I'll call "choice nonsense". Here's an example:
Given two sets of points $A$ and $B$ in the plane, the distance from $A$ to $B$ is defined as follows: pick a point $(x, y)$ in $A$ and a point $(p, q)$ in $B$; the distance is then $\sqrt{(x-p)^2 + (y-q)^2}$.
The problem here is that if you chose a different point in $A$ or $B$, you'd get a different value for the "distance" -- the apparent "definition" is horribly ambiguous. (It's also problematic if either $A$ or $B$ is empty, of course.)
OK, back to addition in modular arithmetic, I'm going to skip the subscript $m$ in what follows, and just use square brackets to keep notation less baroque.
Here's a definition of addition for modular arithmetic classes:
To add class $a$ to class $b$, observe that both $a$ and $b$ are nonempty, so $a$ contains some integer $x$ and $b$ contains some integer $y$. Compute $x+y$, and take the residue class $c$ of this number $\bmod m$, and declare $a \oplus b = c$.
That is a more process-oriented description of the short form, $[x] \oplus [y] = [x + y]$. And in its more verbose form, it reveals that there's a potential choice-ambiguity in the "definition".
The trick here is to prove a lemma:
Lemma: if $x \equiv x' \bmod m$ and $y \equiv y' \bmod m$, then $x+y \equiv x' + y' \bmod m$.
Now you can look at the definition and see that if you and I each choose representatives of the classes $a$ and $b$ (you chose $x$ and $y$, I choose $x'$ and $y'$), we'll both get the same residue class $c$ using the definition. The apparent ambiguity is actually not an ambiguity at all.
Let me give an alternative (worse...see below) definition of addition of classes in modular arithmetic. We define, for classes $a$ and $b$, that $$ a \oplus b = \{ u + v \mid u \in a, v \in b \}. $$ In other words: we simply add every number in $a$ to every number in $b$. This definition has the advantage that there's no choice-ambiguity at all: it explicitly constructs a set.
There's still a challenge here, though: as defined, $a \oplus b$ is certainly a set of integers...but is it a residue class $\bmod m$? You'd need to check that: (i) any two elements of $a \oplus b$, as defined, differ by a multiple of $m$, and (ii) that for any element $s$ of $a \oplus b$, all numbers of the form $s + km$ are also in $a \oplus b$. Both of those happen to be true and no harder to prove than my lemma.
Now what happens if we try to define $a \otimes b$ in the same way? We (tentatively) define $$ a \otimes b = \{ u v \mid u \in a, v \in b \}. $$
It'll turn out that item i still holds just fine (another small lemma), but item ii is generally false. And that's why the alternate definition isn't as good, despite avoiding choice-ambiguity.
I find it's useful to draw a diagram:
$$ \require{AMScd} \newcommand {Z} {{\Bbb Z}} $$ \begin{CD} \Z \times \Z @>{\cdot}>> Z;\\ @VVV @VVV \\ \Z/m\Z \times \Z/m\Z @>{\otimes}>> \Z \end{CD}
where the vertical maps are "reduce mod $m$". This diagram should be "commutative", i.e., if you follow any two arrows (which both represent homomorphisms) from the upper left to the lower right, you should get the same thing. That's the goal we aspire to if we're going to define $\otimes$ correctly --- it's kind of a visual representation of the question you asked.
What the usual definition does is to (secretely) invert the left two arrows to produce this, in which the three-arrow sequence is used to define $\otimes$:
\begin{CD} \Z \times \Z @>{\cdot}>> Z;\\ @AAA @VVV \\ \Z/m\Z \times \Z/m\Z @>{\otimes}>> \Z \end{CD}
The problem is that there's isn't a nice homomorphism from $\Z/mZ$ to $\Z$. You have lots of choices when you go from a residue class to an element of that class. So if you want to use that diagram to define $\otimes$, you have to prove that whatever choices you make, you get the same result by following the three arrows.
The proof that you get the same result is exactly the short sketch that @G.F. gave as an answer. (Or, for $\oplus$, the little lemma that I wrote above.)