Question #152148

Define a relation `~' on R by 'a ~ b if a — b is an integer'. Show that this is an equivalence relation. Give the equivalence classes of 0 and √2 .


1
Expert's answer
2020-12-21T17:31:43-0500

Let us show that a relation '\sim' on R\mathbb R by 'aba \sim b if aba - b is an integer' is an equivalence relation, that is reflexive, symmetric and transitive.


For each aRa\in\mathbb R, aa=0Za-a=0\in\mathbb Z , and thus aaa\sim a each aRa\in\mathbb R, and the relation is reflexive.


If aba\sim b, then abZa-b\in\mathbb Z. It follows that ba=(ab)Zb-a=-(a-b)\in\mathbb Z, and thus ba.b\sim a. Consequently, the relation is symmetric.


Let aba\sim b and bcb\sim c. Then abZ, bcZa-b\in\mathbb Z, \ b-c\in\mathbb Z. It follows that ac=(ab)+(bc)Za-c=(a-b)+(b-c)\in\mathbb Z, and thus aca\sim c. Therefore, the relation is transitive.


By defenition, the equivalence class of an element aa is [a]={bR  baZ}[a]=\{b\in\mathbb R\ |\ b-a\in\mathbb Z\}.


Let us give the equivalence classes of 0 and 2\sqrt{2} :


[0]={bR  b0Z}={bR  bZ}=Z,[0]=\{b\in\mathbb R\ |\ b-0\in\mathbb Z\}=\{b\in\mathbb R\ |\ b\in\mathbb Z\}=\mathbb Z,


[2]={bR  b2Z}={bR  b=2+n, nZ}=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.




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