Consider the following relations. R $$=$$ $${(x$$, $$y)$$ | x, y are real numbers and x $$=$$ wy for some rational number $$w}$$ $$S=mn$$,pqm,n,p,q are integer such that n.q ≠ $$0$$ and $$qm=pn}Statement-1$$: S is an equivalence relation but R is not an equivalence relation.$$Statement-2$$: R and S both are symmetric.

Question

Consider the following relations. R $$=$$ $${(x$$, $$y)$$ | x, y are real numbers and x $$=$$ wy for some rational number $$w}$$ $$S=mn$$,pqm,n,p,q are integer such that n.q ≠ $$0$$  and  $$qm=pn}Statement-1$$: S is an equivalence relation but R is not an equivalence relation.$$Statement-2$$: R and S both are symmetric.

Infinity Learn · Accepted Answer

Since (0$$, $$1) ∈ R but (1$$, $$0) ∈ R, R is not symmetric and hence is not an equivalence relation so $$statement-2$$ is false. Next, For the relation S,$$qm=pn$$⇒$$mn=pqThus$$ mn,pq∈S⇒$$mn=pqwhich$$ shows that S is reflexive and symmetricAgain , mn,pq∈S and pq,rs∈S⇒ $$mn=pq=rs$$ ⇒mn,rs∈SThus S is transitive and hence S is an equivalence relation. So, statement $$1$$ is true

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Consider the following relations. R = {(x, y) | x, y are real numbers and x = wy for some rational number w} S=mn,pqm,n,p,q are integer such that n.q ≠ 0 and qm=pn}Statement-1: S is an equivalence relation but R is not an equivalence relation.Statement-2: R and S both are symmetric.

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