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  • Let P(S) denote the power set of <img alt="\mathrm{S}=\{1,2,3, \ldots \ldots \ldots, 10\}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BS%7D%3D%5C%7B1%2C2%2C3%2C%20%5Cldots

Let P(S) denote the power set of \mathrm{S}=\{1,2,3, \ldots \ldots \ldots, 10\} Define the relations \; R_{1} \; and \; R_{2} on P(S) as 

\mathrm{AR}_1 \mathrm{~B} \text { if }\left(\mathrm{A} \cap \mathrm{B}^{\mathrm{C}}\right) \cup\left(\mathrm{B} \cap \mathrm{A}^{\mathrm{C}}\right)=\varnothing and \mathrm{AR}_2 \mathrm{~B} if \mathrm{A} \cup \mathrm{B}^{\mathrm{C}}=\mathrm{B} \cup \mathrm{A}^{\mathrm{C}}, \forall \mathrm{A}, \mathrm{B} \in \mathrm{P}(\mathrm{S}) . Then :

Option: 1

only R_{1} is an equivalence relation


Option: 2

only \mathrm{R}_2 is an equivalence relation


Option: 3

Both \mathrm{R}_1 and \mathrm{R}_2 are equivalence relations


Option: 4

both \mathrm{R}_1 and \mathrm{R}_2 are not equivalence relations


Answers (1)

best_answer

\begin{aligned} & S=\{1,2,3, \ldots \ldots 10\} \\ & P(S)=\text { power set of } S \\ & A R, B \Rightarrow(A \cap \vec{B}) \cup(\vec{A} \cap B)=\phi \end{aligned}

R1 is reflexive, symmetric
For transitive

(\mathrm{A} \cap \vec{B}) \cup(\overrightarrow{\mathrm{A}} \cap \mathrm{B})=\phi ;\{\mathrm{a}\}=\phi=\{\mathrm{b}\} \mathrm{A}=\mathrm{B}$ $(B \cap \vec{C}) \cup(\vec{B} \cap C)=\phi \therefore B=C

\therefore \mathrm{A}=\mathrm{C} \text { equivalence }

\begin{aligned} & \mathrm{R}_2 \equiv \mathrm{A} \cup \vec{B}=\vec{A} \cup \mathrm{B} \\ & \mathrm{R}_2 \rightarrow \text { reflexive, symmetric } \\ & \text { for transitive } \end{aligned}

\begin{aligned} & \mathrm{A} \cup \overline{\mathrm{B}}=\overrightarrow{\mathrm{A}} \cup \mathrm{B} \Rightarrow\{\mathrm{a}, \mathrm{c}, \mathrm{d}\}=\{\mathrm{b}, \mathrm{c}, \mathrm{d}\} \\ & \{\mathrm{a}\}=\{\mathrm{b}\} \therefore \mathrm{A}=\mathrm{B} \\ & \mathrm{B} \cup \overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{B}} \cup \mathrm{C} \Rightarrow \mathrm{B}=\mathrm{C} \end{aligned}

\therefore \mathrm{A}=\mathrm{C} \quad \therefore \mathrm{A} \cup \overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{A}} \cup \mathrm{C} \therefore \text { Equivalence }

 

Posted by

manish painkra

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