For x, y ∈ R we write x ∼ y if x - y is an integer. a) Show that ∼ is an equivalence relation on R. b) Show that the set [0, 1) = {x ∈ R : 0 ≤ x < 1} is a set of representatives for the set of equivalence classes. More precisely, show that the map Φ sending x ∈ [0, 1) to the equivalence class C (x) is a bijection.

A. It is an equivalence relation on R

B. In fact, the set [0,1) is a set of representatives

Step-by-step explanation:

A. The definition of an equivalence relation demands 3 things:

The relation being reflexive (∀a∈R, a∼a) The relation being symmetric (∀a, b∈R, a∼b⇒b∼a) The relation being transitive (∀a, b, c∈R, a∼b^b∼c⇒a∼c)

And the relation ∼ fills every condition.

∼ is Reflexive:

Let a ∈ R

it's known that a-a=0 and because 0 is an integer

a∼a, ∀a ∈ R.

∼ is Reflexive by definition

∼ is Symmetric:

Let a, b ∈ R and suppose a∼b

a∼b ⇒ a-b=k, k ∈ Z

b-a=-k, - k ∈ Z

b∼a, ∀a, b ∈ R

∼ is Symmetric by definition

∼ is Transitive:

Let a, b, c ∈ R and suppose a∼b and b∼c

a-b=k and b-c=l, with k, l ∈ Z

(a-b) + (b-c) = k+l

a-c=k+l with k+l ∈ Z

a∼c, ∀a, b, c ∈ R

∼ is Transitive by definition

We've shown that ∼ is an equivalence relation on R.

B. Now we have to show that there's a bijection from [0,1) to the set of all equivalence classes (C) in the relation ∼.

Let F: [0,1) ⇒ C a function that goes as follows: F (x) = [x] where [x] is the class of x.

Now we have to prove that this function F is injective (∀x, y∈[0,1), F (x) = F (y) ⇒ x=y) and surjective (∀b∈C, Exist x such that F (x) = b):

F is injective:

let x, y ∈ [0,1) and suppose F (x) = F (y)

[x]=[y]

x ∈ [y]

x-y=k, k ∈ Z

x=k+y

because x, y ∈ [0,1), then k must be 0. If it isn't, then x ∉ [0,1) and then we would have a contradiction

x=y, ∀x, y ∈ [0,1)

F is injective by definition

F is surjective:

Let b ∈ R, let's find x such as x ∈ [0,1) and F (x) = [b]

Let c=║b║, in other words the whole part of b (c ∈ Z)

Set r as b-c (let r be the decimal part of b)

r=b-c and r ∈ [0,1)

Let's show that r∼b

r=b-c ⇒ c=b-r and because c ∈ Z

r∼b

[r]=[b]

F (r) = [b]

∼ is surjective

Then F maps [0,1) into C, i. e [0,1) is a set of representatives for the set of the equivalence classes.