Equivalence relation and equivalence class I have looked around the internet for an easy to approach - down to earth explanation of equivalence relation & equivalence class but having no success. If any of you can explain in very basic term please tell me + give an example. Thanks in advance.
 A: Think of classifying something like cars. Describe all cars with one characteristic - say color. Then, although there are zillion of cars in the world, there are only a few different colors of cars. Each color is an equivalence class. So now instead of the list of all the cars starting with "mine" and ending with "that last car I haven't counted", you can count all the cars in the world by listing all the colors that cars come in. Which is an easier classification, but contains less information.
This simple example doesn't try to incorporate any extra structure - cars is a set, not a ring or a group or something else.
A: Think of all the integers... now divide them all by four, for example.  
There are integers that will have a remainder of $0$, like $4, 8, 12, 16,$ etc.
There are integers that will have a remainder of $1$, like $1, 5, 9, 13, $ etc.  
There are integers that will have a remainder of $2$, like $2, 6, 10, 14,$ etc.
There are integers that will have a remainder of $3$, like $3, 7, 11, 15,$ etc.
Numbers divisible by 4 go into the class $$[0]=\{-4,0,4,8,12,...\}$$
Numbers not divisible by 4 go into their own distinct equivalence classes.  So the integers that leave a remainder of 1 are in their own class $$[1] = \{...,-3,1,5,9,13,...\}$$ integers that leave a remainder of 2 are in their own class $$[2] = \{...,-2,2,6,10,14,...\}$$ integers that leave a remainder of 3 are in their own class $$[3]=\{...,-1,3,7,11,15,...\}$$  
Thus you can see all the integers find a home somewhere based upon this divisibility relation.  
This is a math example, but I think it is simple enough to see the relation work and form the equivalence classes.
EDIT (for Peter who asked me to expound on properties of equivalence relations; Also, this is where it will definitely get a little Mathy!!!):
Equivalence relations have three properties; reflexivity, symmetry, and transitivity.  Now in our case, remember, we are relating all numbers with their remainder when dividing by 4.  So when I am looking at members of each equivalence, I am looking for the divisibility relation.  So two numbers $a, b$ are related from a particular class if
$$a-b \text{ is divisible by } 4.  \text{(or mathematically } 4|(a-b))$$
I don't think this is a stretch to see.  Just look at all the elements of each class...they are all separated by 4.  In $[2]$ for example, 
$$14-6=8=4\cdot{2}$$
So this is how we will actually compare elements for equivalence.  Now our three properties we need to satisfy are;
1)  Reflexivity; This just means a number is related to itself.  In our case, let's look at the element $5$ from the class $[1]$.  Is $5$ related to itself (seems intuitively to be the case right?).  Apply the relation...
$$5-5=0=4\cdot{0}$$
$0$ is certainly divisible by $4$, so our relation is Reflexive...if we let our element be just $a$, I think it's clear it is generally true for all integers...
2)  Symmetry; this just means if one number can be related to a second, it is the same as if we relate the second to the first... So let's stay in class $[1]$ and relate $5$ and $9$.  Apply the relation...
$$5-9=-4=4\cdot{(-1)}$$
now multiply both sides by $-1$ and see that
$$(-1)(5-9)=9-5=4=4\cdot{1}=(-1)4\cdot{(-1)}$$
In the middle we can see that we have our statement we need to show symmetry.  If $5$ is related to $9$, then $9$ is also related to $5$ (since $9-5$ is divisible by 4 also...).  In generalized terms, if $a$ is related to $b$, then $b$ is related to $a$.
3)  Transitivity; this is the old standard; if $a$ is related to $b$ and if $b$ is related to $c$, then $a$ is also related to $c$.  Lets finish in $[1]$.  Let my three numbers be $13,5,-3$.  Apply the relation on the first two...
$$13-5=8=4\cdot{2}$$
Apply the relation on the last two
$$5-(-3)=8=4\cdot{2}$$
Then
$$13-5+5-(-3)=13+3=16=4\cdot{4}$$
Therefore it is Transitive.  
