For a given relation in $\mathbb{N}\times \mathbb{N}$ find the number of elements in it's equivalence class

Question

The whole problem goes like this:

We define the relation $R$ in $\mathbb{N}\times \mathbb{N}$ in the following way:

$(a,b)R(c,d)$ iff $a-d=c-b$

First find proof the it's a relation of equivalence (done). Then, and here i have a problem, for each $(a,b)$ belonging to $\mathbb{N}\times \mathbb{N}$ find the amount of elements in it's equivalence class.

Any help will do.

Thanks.

Answer 1

Let's say $(a, b) \in \Bbb{N} \times \Bbb{N}$. We want to find the set of all $(c, d) \in \Bbb{N} \times \Bbb{N}$ such that $a-d=c-b$. This equation can be rearranged to say $a+b=c+d$. Thus, the equivalence class of $(a, b)$ is the following: $$\{(c, d) \in \Bbb{N} \times \Bbb{N} \mid a+b=c+d\}.$$

For any constant $n \in \mathbb N$, there are precisely $n$ elements $(c, d) \in \mathbb N \times \mathbb N$ with the property that $c + d = n$: $$(0, n); (1, n-1); (2, n-2); (3, n-3); \dotsc; (n, 0).$$

Applying this to $n = a + b$, we see that the equivalence class of $(a, b)$ has $a+b$ elements in it.