Show equivalence class of equality How do I show equivalence class of "="
let R={(a,a) such that $a\in$ A}. 
I know I have to show reflexive, transitive, symmetry properties to show that this relation is an equivalence relation. 


*

*Reflexive: For all a in A, a=a. proved.

*Symmetry: Suppose a=b, then a=a =>b=a. 

*Transitive: Suppose a=b and b=c, then a=b=c. So, a=c. 


But, I have trouble showing its equivalent class of "=". Any help is appreciated. Thank you. 
 A: Since you never have any $(a,b) \in R$ with $a \not = b$, that means that every $a$ forms an equivalence class all by itself for any $a \in A$. Or, if you want: $[a] = \{ a \}$ for every $ a \in A$.
A: Let $(a,b)\in A^2$.


*

*If $(a,b)\in R$, then $a=b$.

*If $a=b$, then $(a,b)\in R$.
Therefore, $(a,b)\in R$ if and only if $a=b$. 
Does it help?
A: By definition of $R$, $(a, a) \in R$.  So, $R$ is reflexive.
Suppose $(a, b) \in R$.  By definition, $(a, b)$ must be of the form $(c, c)$ for some $c$.  But then $(c, a) \in R$ by the indiscernibility of identicals (substitutivity of equals).  Similarly, $(b, a) \in R$ by a second application of the same principle.  So, if $(a,b) \in R$, $(b, a) \in R$.  That is, $R$ is symmetric.
Suppose $(a,b) \in R$ and $(b,c) \in R$. By definition, $(a,b)$ must be of the form $(d,d)$ for some $d$.  But, then, $(b,c) \in R$ implies that $(d,c) \in R$ by the indiscernibility of identicals (substitutivity of equals).  A second application of the same principle yields $(a,c) \in R$.  So, $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$.  That is, R is transitive.
Since $R$ is reflexive, symmetric, and transitive, $R$ is an equivalence relation.  
This is one of those nasty little bootstrap proofs that seem to make no sense unless one understands that one is relating two distinct notations to one another.  The sign of equality "warrants" substitutions in the notation of the class elements as ordered pairs.
And, I just noticed -- proving that a relation is an equivalence relation is not "showing the equivalence class" of a relation.  I think your wording is leading you to some confusion.
