Having trouble with this question; have what I think is an answer but hoping for

Question

Having trouble with this question; have what I think is an answer but hoping for a second opinion. There is a relation T on RxR (i.e. cartesian plane with R being all real numbers), with the rule (a,b) T (c,d) if and only if (a-b) = (c-d). Prove that this is an equivalence relation and describe the equivalence classes.

I think this is reflexive, symmetric, and transitive because we are dealing with an infinite possibility for (a,b), and can build a reflexive/symmetric/transitive counterpart (c,d) as a result. In other words, if (a,b) is (5,4) and (6,3), there will be some other (c,d) that is (5,4) and (6,3) satisfying another (a,b) condition to make it reflexive, another (c,d) that is (4,5), (3,6) satisfying another (a,b) condition to make it symmetric, and likewise an equivalent transitive counterpart.

Description of equivalence classes; for any ordered pair (a,b), there is an equivalent (c,d) that is equal to (a,b) for reflexive, (b,a) for symmetric, and (a+y, b+y) for transitive.

Does this make sense or am I completely off?

Explanation / Answer

Reflexive means (a,b)T(a,b) is true.

Here a-b = a-b Hence it is reflexive.

Symmetric:

(a,b) T (c,d) means a-b = c-d

This proves that c-d = a-b or (a,b) T (c,d)

To prove transitive let (a,b) T (c,d) and (c,d) T (e,f) are true.

Then a-b = c-d,   Also c-d = e-f,.It follows that a-b = e-f. Thus transitive.

As it is symmetric, reflexive and transitive, T is an equivalence relation.

This is how we have to prove it.

Infinitely many like that you need not prove.

This is the way to prove a relation is an equivalence one.

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