In the set of the integers Z, given a positive number m, we define $ \sim = \lbrace (x1,x2) | x1-x2 = m z, z \in Z \rbrace $ Proof that ~ is an equivalence relation. How many equivalence classes does it have in Z?
Here I need to proof that ~ is reflexive, symmetric and transitive. Reflexive: I need to proof that given a number x we always have: $$ x \sim x $$ so $ x - x = m z $ the solution is $ z = 0 \in Z$ so the relation is reflexive.
Symmetric: $ x \sim y $ implies that $ y \sim x $ so: $ x - y = m z $ and $ y - x = m z $ which can be written as $ x - y = - m z $ but the existance of a $ z \in Z$ that verifies the condition depends from the value of m right? Because if z were rational then I will always find a z but here it is an integer, so if for example $ x = 5, y = 2$ I get $ 5 - 2 = m z$ so $ 3 = mz$ but here z should be $ z = 3/m$ which is not always an integer.
Transitive:
$ x \sim y $ and $ y \sim c $ implies that $ x \sim c $ so:
$ x - y = m z $ and $ y - c = m z $ which can be written as $ y = m z + c $ so substituting in the first I get
$ x - mz - c = m z $ thus $ x - c = 2m z $ where I can find a $ z \in Z$ ? (the problem is the same as was before).
Now how many equivalence classes does it have?
There are total of $m$ equivalence classes, and they are: $[0], [1], ..., [m-1]$.