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14 June, 14:12

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.

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  1. 14 June, 14:18
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    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.
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