Answer:
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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