Equivalence class in equivalence relation

Question

Let $A=\{1,2,3,4,...,9\}$ and $R$ be a relation in $A \times A$ defined by $(a,b)R(c,d)$. If $a+d=b+c$ for $(a,b),(c,d)$ in $A \times A$. Prove that $R$ is an equivalence relation and also obtain the equivalent class $(2,5)$.

I have proved it to be equivalence relation but how to find the obtain the equivalent class $(2,5)$. Could someone help me with this?

Answer 1

You know the theorem that if $A$ is a set and $\sim$ is an equivalence relation on $A$, then $A$ can be split into disjoint classes under this equivalence.

The key to this theorem is the following observation:

Two elements in the same equivalence class are related under $\sim$, while elements of different classes are not related under $\sim$.

That is, given an element, it's equivalence class is precisely the set of all elements it is related to.

Having said that, you just need to figure out all pairs of numbers $(a,b) \subset A \times A$ such that $a - b = 2-5$, with the help of the comment given above. I am sure you can do it from here.