A proof about equivalence relation and classes 
Let $R$ be an equivalence relation in the set $A$ and $a,b \in A$. Show that $R(a)=R(b)$ iff $aRb$.

We must prove two implications here. First, that $$R(a)=R(b) \Rightarrow aRb$$
Note that $R(a):= \{b\in A : aRb\}$, and $R(b):=\{a\in A : bRa\}$. Also since these sets are equal, they must be subsets of each other $$R(a) \subseteq R(b)$$
It follows that $\forall b \in A$ : $$b\in R(a) \Rightarrow b \in R(b)$$
If $b$ is an element of both of these equivalence classes, then certainly $b=a$. From the fact that $$R(b)\subseteq R(a)$$
it follows similarly, that $a=b$. I'm pretty sure this shows
 $aRb$, but I'm not sure how I should word it. As for $$aRb \Rightarrow R(a)=R(b)$$
Since $R$ is an equivalence relation, we have by symmetry that, $\forall a,\forall b \in A$ $$aRb \Rightarrow bRa$$
This is where I am at the moment. I just wanted to check if the proof is ok up to this point.
 A: 
If b is an element of both of these equivalency classes, then certainly b=a

I can not see how you think that.  That makes no sense.  Unless this class has only one element, then (because they are equal) all the elements are in both and all the elements can't all be equal to each other.
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Use the definition of equivalence.
So to prove $R(a)=R(b)\to aRb$.....
If $R(a) = \{x\in A: aRx\}$....(I advice not using the variable "$b$" because you use that variable for another element... oh, maybe that caused your confusion....) 
then as $R$ is an equivalence class then by reflexivity $aRa$ and so $a \in R(a)$.
But if $R(a) = R(b) = \{x\in A: bRx\}$ it follows that $a\in R(b)$ so $bRa$. 
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Can you finish this?

 but symmetry as $bRa$ we know $aRb$.

And to prove $aRb \to R(a)=R(b)$.

 If $c\in R(a)$ then $cRa$ but $aRb$ so .......

.........

 .... by transitivity $cRb$ so ...

.......

  ..... $c \in R(b)$ and $R(a)\subset R(b)$.  If $c \in R(b)$ then .....

......

  ...... $cRb$.  But $aRb$ so by symmetry $bRa$ and by transitivity $cRa$ and be symmettry $aRc$ so $c\in R(a)$ so $R(b)\subset R(a)$ and so $R(a) = R(b)$.

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It may help to do an example.  $\equiv \pmod 5$ is an equivalence relation.
So $[a] = \{x\in \mathbb Z| x\equiv a \pmod 5\}= \{a + 5k|k\in \mathbb N\}$ and $[b] = \{x\in \mathbb Z| x\equiv a \pmod 5\}= \{b + 5k|k\in \mathbb N\}$.
the does  $[a] = [b]$ mean that $b\equiv a \pmod 5$?
Well, yes....
$a \equiv a \pmod 5$ so $a \in [a] = [b]$ so $a\equiv b\pmod 5$ so $b \equiv a\pmod 5$.
If if $a \equiv b\pmod 5$ does that mean $[a] = [b]$?
Well, yes.....
If $a \equiv \pmod 5$ then there is a $k$ so that $a = b + 5k$.
Then $c\in [a]\iff c \equiv a \pmod 5\iff $ there is a $j$ so that $c = a+5j\iff c=(b+5k)+5j=b+5(k+j)\iff c\equiv b\pmod 5\iff c\in [b]$.
So $[a]$ and $[b]$ have the exact same elements.
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In general. 
Suppose $R(a) = R(b)$. then $bRb\implies b\in R(b)\implies b\in R(a)\implies aRb$.
Suppose $aRb$. then $c\in R(a)\implies aRc\implies cRa\implies cRa$ and $aRb\implies cRb \implies c\in R(b)\implies cRb\implies cRb$ and $ aRb\implies cRb$ and $bRa\implies cRa\implies c\in R(a)$.
So $c\in R(a)\iff c\in R(b)$ and so $R(a) = R(b)$.
That's all there is to it. 
A: If $aRb$, then $x \in R(a)$ implies that $xRa$ since $R$ is symmetric so by transivity $xRb$ and $x \in R(b)$. The other inclusion is similar. We know that $bRb$ by reflexivity so that if $R(a) = R(b)$, then $b \in R(a$) and $aRb$.
A: I would do something like that.
Suppose first that $R(a)=R(b)$.
As $R$ is reflexive, we have $aRa$ which implies $a\in R(a)$. By hypothesis, we also have $a \in R(b)$, i.e $aRb$.
Conversely let’s suppose that $aRb$ and take $c \in R(a)$, which means $cRa$. By transitivity, we also have  $cRb$, i.e. $c \in R(b)$. Therefore, we have proven that  $R(a)\subseteq R(b)$. Now, by symmetry if we suppose $aRb$, we also have $bRa$. With a similar proof that what we just did, we can conclude that $R(b) \subseteq R(a)$. Finally $R(a)=R(b)$.
Overall, I would suggest that you use more explicitly the properties of an equivalence relation. 
