Let S be the set of real numbers. If a,b are element of S, define a ~ b if a - b is an integer. Show that ~ is

Question

an equivalence relation on S. Decribe the equivalence classes of S

Answer 1

a~a, because a-a =0 which is an integer
a~b, means that a-b =n an integer, now multiply both sides by -1,
and we get b-a = -n which is also an integer => b~a
a~b, b~c => a-b =n and b-c =m, now add them up
and we get: a-b +b-c = a-c = n+m which is an integer => a~c .

The equivalence classes are [0,1) . Notice that every number in the interval is related to number outside the interval, for example
1/2~ 3/2~ -1/2 ~ 5/2, etc.
or 0~1~2 ... etc
and so every number in the interval [0,1) can be considered a representative of an equivalence class.

alternatively, we can think about it like this:
[0,1]/~ = interval with 0 and 1 identified, i.e.
[0,1]/~ = circle .

Answer 2

Has to satisfy all three of:
reflexivity
symmetry
transitivity


So for reflexivity: let a be a real number. a ~ a since a- a = 0, which is an integer, so it's reflexive.
Symmetry: Let a ~ b. We need to show b ~ a is also there. But if a - b is an integer, then b - a = - (a-b) is also an integer. Therefore we have symmetry.
Trans: Assume a ~b, and b~c. We need to show a~c. But a - b and b-c are integers, and adding one to the other gives a-b+b-c = a-c, which must also be an integer since each of a,b,and c are integers.

Therefore all three parts are satisfied, so this is an equiv. rel.