The problem is
Let S be the set of integer and let R be the set of pairs $(a,b)$ such that $a+b$ is even. Describe the equivalence classes.
My solution was $a + b = k$ for some even integer $k.$ Therefore, $a = k - b$ for some even integer $k.$
$[a] = \{k - b \mid k \in \mathbb Z\}$.
Also, $b = k - a$ for some even integer $k.$
$[b] = \{k - a \mid k \in \mathbb Z\}$.
I don't know whether or not my solution is correct. I personally think my solution is missing something but don't know what to add. Please help me with this problem. Thank you.
"Let S be the set of integers and let R be the set of pairs $(a,b)$ such that $a+b$ is even. Describe the equivalence classes."
$(1)$ All elements in $\{\ldots, - 4, -2, 0, 2, 4, 6, \ldots\}$ (i.e., all even integers) are in one equivalence class, because when we pick any two even integers, their sum is even.
All elements in $\{\ldots, -3, -1, 1, 3, 5, \ldots\}$ (i.e., all odd integers) are in another equivalence class, because when we pick any two odd integers, their sum is even.
The two equivalence classes partition the set of integers, $\mathbb Z$.
To be more precise, we'll show that all even numbers are related to one another, and all odd numbers are related to one another, and that no even/odd pair is related to one another, under the given equivalence relation:
Let's take $a$ to be any even integer. So $a = 2k$ for some integer $k$. When we add any other even integer to $a$, say we add $x = 2m$, then we get $a + x = (2k + 2m) = 2(k+m)$ which is again even. So adding two even integers results in an even integer.
Let's take $b$ to be any odd integer. So $b = 2k+1$ for some integer $k$. Suppose we add another odd integer to $b$, say $y = 2n+ 1$. When we add $b + y = 2k+1 + 2n + 1 = 2 k + 2n +2 = 2(k+n+1)$, and hence the sum of any two odd integers is always even.
To show the two equivalence classes are disjoint, let us take any even integer $c = 2p$ and any odd integer $d = 2q+1$ where p and q are integers, and add them: $c + d = 2p + (2q + 1) = 2(p+q) +1$, which is odd, and so it is never the case that an even integer c added to an odd integer d, is in the relation. Hence the set of all even integers, and the set of all odd integers, comprise the two equivalence classes, and their union is $\mathbb Z$, itself.