find whether the relation, defined on set R, the set of real no. defined as S={ (a,b): a,b belongs to R and a-b+root3 belongs to T, the set of irrational no., is reflexive, symmetric and trasitive. can we say that relation S is not an equivalancerelation?

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Our} \mathrm{relation} \mathrm{is} \mathrm{given} \mathrm{by}: \\ \mathrm{aRb} : \mathrm{a}-\mathrm{b}+\sqrt{3} \mathrm{is} \mathrm{irrrational} \\ \left(1\right) \mathrm{Reflexive}: \\ \mathrm{aRa}: \mathrm{a}-\mathrm{a}+\sqrt{3}=\sqrt{3} \mathrm{which} \mathrm{is} \mathrm{irrational}. \\ \mathrm{Hence} \mathrm{aRa} \mathrm{is} \mathrm{true}. \\ \mathbf{\Rightarrow }\mathbf{R}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{reflexive}\mathbf{ }\mathbf{relation}\mathbf{.} \\ \left(2\right) \mathrm{Symmetric} \\ \mathrm{bRa}: \mathrm{b}-\mathrm{a}+\sqrt{3} \\ \mathrm{Now} \mathrm{consider} \mathrm{a} \mathrm{case} \mathrm{when} \mathrm{a}-\mathrm{b}=\sqrt{3}, \mathrm{then} \\ \mathrm{b}-\mathrm{a}=-\sqrt{3} \\ \mathrm{bRa}: -\sqrt{3}+\sqrt{3}=0 \mathrm{which} \mathrm{is} \mathrm{rational}. \\ \mathrm{Hence} \mathrm{bRa} \mathrm{does} \mathrm{not} \mathrm{hold} \mathrm{for} \mathrm{all} \left(\mathrm{a},\mathrm{b}\right) \\ \mathbf{\Rightarrow }\mathbf{R}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{not}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{symmetric}\mathbf{ }\mathbf{relation}\mathbf{.} \\ \left(3\right) \mathrm{Transitive} \\ \mathrm{Let} \mathrm{bRc}: \mathrm{b}-\mathrm{c}+\sqrt{3} \mathrm{is} \mathrm{irrational} \\ \mathrm{Let} \mathrm{b}=2\sqrt{3} \mathrm{and} \mathrm{c}=\sqrt{3} \mathrm{then} \mathrm{above} \mathrm{statement} \mathrm{is} \mathrm{true}. \\ \mathrm{aRc}: \mathrm{a}-\mathrm{c}+\sqrt{3}=\mathrm{a}-\sqrt{3}+\sqrt{3}=\mathrm{a}, \\ \mathrm{As} \mathrm{a}\in \mathrm{R}, \mathrm{hence} \mathrm{a} \mathrm{can} \mathrm{be} \mathrm{rational} \mathrm{or} \mathrm{irrational}. \\ \Rightarrow \mathrm{aRc}: \mathrm{a}-\mathrm{c}+\sqrt{3} \mathrm{is} \mathrm{irrational} \mathrm{will} \mathrm{not} \mathrm{always} \mathrm{be} \mathrm{true}. \\ \mathbf{\Rightarrow }\mathbf{R}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{not}\mathbf{ }\mathbf{a}\mathbf{ }\mathbf{transitive}\mathbf{ }\mathbf{relation}\mathbf{.} \\ \mathrm{As} \mathrm{R} \mathrm{is} \mathrm{neither} \mathrm{symmetric} \mathrm{nor} \mathrm{transitive}. \\ \mathbf{Hence}\mathbf{ }\mathbf{we}\mathbf{ }\mathbf{can}\mathbf{ }\mathbf{say}\mathbf{ }\mathbf{that}\mathbf{ }\mathbf{R}\mathbf{\hspace{0.17em}}\mathbf{is}\mathbf{ }\mathbf{not}\mathbf{ }\mathbf{an}\mathbf{ }\mathbf{equivalence}\mathbf{ }\mathbf{relation}\mathbf{.} \end{gathered}$$

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