What does quotienting by a congruence mean? I have come across quotient algebras in my different mathematics courses. 
I know of quotienting with normal groups, quotienting with ideals etc.
While studying Boolean Algebra I encounter quotienting with congruence relation defined by double equivalence. 
This led to me think what do quotient algebra hold ? I am not sure whether I am able to convey my question. I dont seem to really get "feel" of quotient algebra.
Why is it necesaary for relation to be congruence? Rather why doesn't just equivalence do the trick ?
Also kindly give reference with example of booloean algebras.   
 A: The idea of congruence arises in any algebraic structure that can be studied within the universal algebra setting (this includes groups, rings, boolean algebras). 
Let $A$ be a set and let $\{f_{1},...,f_{m}\} = \mathscr{F}$ be a collection of $n_{i}$-ary functions $A^{n_{i}} \rightarrow A$ where $1 \leq i \leq m$ (so $n_{i}$ depends on $i$). A $0$-ary (nullary) function picks out a fixed element of $A$. $A$ is then an "algebra" over the operations $f_{1},...,f_{m}$ of type $\mathscr{F}$.
Let $\sim$ be an equivalence relation on $A$. We say that $\sim$ is a congruence on $A$ if for every operation $f_{i} \in \mathscr{F}$, we have that if $a_{j},b_{j} \in A$ such that $a_{j} \sim b_{j} \:\forall 1 \leq j \leq n_{i}$ then$$ f_{i}(a_{1},...,a_{n_{i}}) \sim f_{i}(b_{1},...,b_{n_{i}}) \qquad (\star)$$
Then, denoting $A/\sim$ to be the set of $\sim$-classes of $A$, $A/\sim$ is an algebra of type $\mathscr{F}$, because by $(\star)$, $f_{i}([a_{1}],...,[a_{n_{i}}])$ is well-defined.
To check that these ideas coincide with your intuitions about normal subgroups, first check that the following definition of a group in the universal algebra setting coincides with the more familiar definition:
A group is a set $G$ with one binary operation, $\cdot$, one unary operation $^{-1}$ and one nullary operation $1$ satisfying:
\begin{array}{cccc} & (a \cdot b) \cdot c  &=& a \cdot (b \cdot c) \\ & a \cdot a^{-1}&=&a^{-1} \cdot a \\ &a \cdot 1 &=& 1 \cdot a &=& a\end{array}
$\forall a,b,c \in G$. Show that for a congruence $\sim$ of $G$, the $\sim$-class of $1$ is a normal subgroup. Then prove the converse, i.e. that given any normal subgroup $N$ of $G$ you can construct a congruence $\sim$ on $G$ such that $N$ is the $\sim$-class of $1$. What is $G/\sim$?
If this interests you then look at the following book, freely available online: http://www.math.uwaterloo.ca/~snburris/htdocs/UALG/univ-algebra2012.pdf 
Universal algebra generalizes many notions common to many algebraic structures such as the homomorphism theorems, and whilst it does not permit some algebraic structures in the theory (fields for instance), it is nonetheless a great way to get a clearer, more comprehensive and general feel for algebra.
