Let us show that a relation '∼ \sim ∼ ' on R \mathbb R R by 'a ∼ b a \sim b a ∼ b if a − b a - b a − b is an integer' is an equivalence relation, that is reflexive, symmetric and transitive.
For each a ∈ R a\in\mathbb R a ∈ R , a − a = 0 ∈ Z a-a=0\in\mathbb Z a − a = 0 ∈ Z , and thus a ∼ a a\sim a a ∼ a each a ∈ R a\in\mathbb R a ∈ R , and the relation is reflexive.
If a ∼ b a\sim b a ∼ b , then a − b ∈ Z a-b\in\mathbb Z a − b ∈ Z . It follows that b − a = − ( a − b ) ∈ Z b-a=-(a-b)\in\mathbb Z b − a = − ( a − b ) ∈ Z , and thus b ∼ a . b\sim a. b ∼ a . Consequently, the relation is symmetric.
Let a ∼ b a\sim b a ∼ b and b ∼ c b\sim c b ∼ c . Then a − b ∈ Z , b − c ∈ Z a-b\in\mathbb Z, \ b-c\in\mathbb Z a − b ∈ Z , b − c ∈ Z . It follows that a − c = ( a − b ) + ( b − c ) ∈ Z a-c=(a-b)+(b-c)\in\mathbb Z a − c = ( a − b ) + ( b − c ) ∈ Z , and thus a ∼ c a\sim c a ∼ c . Therefore, the relation is transitive.
By defenition, the equivalence class of an element a a a is [ a ] = { b ∈ R ∣ b − a ∈ Z } [a]=\{b\in\mathbb R\ |\ b-a\in\mathbb Z\} [ a ] = { b ∈ R ∣ b − a ∈ Z } .
Let us give the equivalence classes of 0 and 2 \sqrt{2} 2 :
[ 0 ] = { b ∈ R ∣ b − 0 ∈ Z } = { b ∈ R ∣ b ∈ Z } = Z , [0]=\{b\in\mathbb R\ |\ b-0\in\mathbb Z\}=\{b\in\mathbb R\ |\ b\in\mathbb Z\}=\mathbb Z, [ 0 ] = { b ∈ R ∣ b − 0 ∈ Z } = { b ∈ R ∣ b ∈ Z } = Z ,
[ 2 ] = { b ∈ R ∣ b − 2 ∈ Z } = { b ∈ R ∣ b = 2 + n , n ∈ Z } = 2 + Z . [\sqrt{2}]=\{b\in\mathbb R\ |\ b-\sqrt{2}\in\mathbb Z\}=\{b\in\mathbb R\ |\ b=\sqrt{2}+n,\ n\in\mathbb Z\}=\sqrt{2}+\mathbb Z. [ 2 ] = { b ∈ R ∣ b − 2 ∈ Z } = { b ∈ R ∣ b = 2 + n , n ∈ Z } = 2 + Z .
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