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Let R be a relation on the set Q of all ...

Let R be a relation on the set Q of all rationals defined by `R={(a,b):a,binQ" and "a-binZ}.` Show that R is an equivalence relation.

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Given:`R={(a,b):a,binQ" and "a-binZ}.`
(i) Let `a inQ.` Then, `a-a=0 inZ.`
`:." "(a,a)inR" for all "a inQ.`
So, R is reflexive.
(ii) `(a,b) inRimplies(a-b)InZ,` i.e., `(a-b)` is an integer
`implies-(a-b)` is an integer
`implies" "(a-c)` is an integer
`implies" "(a,c)inR.`
Thus, `(a,b)inRimplies(b,a)inR.`
`:." "` R is symmetric.
(iii) `(a,b)inR" and "(b,c)inR`
`implies" "(a-b)" is an integer and "(b-c)" is an integer"`
`implies" "{(a-b)+(b-c)}" is an integer"`
`implies" "(a-c)" is an integer"`
`implies" "(a,c)inR.`
Thus, `(a,b)inR" and "(b,c)inRimplies(a,c)inR.`
`:." "` R is transitive.
Thus, R is reflexive, symmetric and transitive.
So, R is an equivalence relation.
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