Number of classes an equivalence relation partitions a set X I am struggling a bit with regards to sets and relations and would appreciate any help.
The question goes as follows:
Consider the relations R and S, defined on the set X = {1, 2, . . . , 99} as follows.

xRy ⇐⇒ x + y is a multiple of 11,
xSy ⇐⇒ x − y is a multiple of 11.

Determine which of those two is an equivalence relation , and find out into 
how many classes does it partition the set X?

I know the definition of equivalence relation ,but it seems to me that both those relations are not reflexive. For R there can be (1,10) and for S there can be (99,88).
As for the second part, I have no idea whatsoever what classes and partitions mean as the given definition is quite vague.
 A: HINT: Is $x+x$ divisible by 11 for integer $x$? And what about $x-x$? (The division in your question is in integers, not naturals).
The classes are $\{1,1+11,1+22,1+33,\ldots,1+88\}$, $\{2,2+11,2+22,\ldots,2+88\}$, $\ldots$
Can you finish this?
A: For equivalence relation, we need to check three criteria: Reflexivity, symmetry, and transitivity.
For reflexivity, you are verifying that $xRx$ and $xSx$ for all $x \in X$. As GNU Supporter suggested, $1R1$ will fail because $1+1=2$ is not a multiple of 11. We do not need to check symmetry or transitivity for $R$ because if it is not reflexive, it is not an equivalence relation.
To see that $xSx$ is true, $x-x=0=0\cdot 11$ is clearly a multiple of 11 for any $x$.
For symmetry, you need $xSy$ implies $ySx$. Hint, if $xSy$, then there exists an integer $k$ such that $x-y = 11k$. If $k$ is an integer, then $-k$ is as well. Use that to show that $S$ is symmetric.
For transitivity, assume $xSy$ and $ySz$ for some $x,y,z \in X$. This means there exists integers $k,m$ such that $x-y = 11k$ and $y-z = 11m$. Add the two equations and on the LHS, you get $x-z$. What do you get on the RHS? Can you show this implies transitivity?
Przemyslaw already explained what the equivalence classes are. A partition of the set means that every element will belong to some equivalence class, and no element will be in multiple classes. The notation for the equivalence class of a certain number is adding brackets around it. So $[1]$ is the equivalence class of $1$. You can write it as $[n] = \{x \in X|nSx\}$. Since the equivalence relation partitions the set, the fact that $1S2$ is false (since $1-2=-1$ is not divisible by 11), then $[1] \cap [2] = \emptyset$. This is how you can partition the set.
