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Let \mathrm{A}=\{0,34,6,7,8,9,10\} and \mathrm{R} be the relation defined on A such that \mathrm{R}=\{\mathrm{x}, \mathrm{y}) \in \mathrm{A} \times \mathrm{A}: \mathrm{x}-\mathrm{y} is odd positive integer or x-y=2\. The minimum number of elements that must be aadded to the relation R, so that it is a symmetric relation, is equal to _____________.

Option: 1

19


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

\begin{aligned} & A=\{0,3,4,6,7,8,9,10\} \quad 3,7,9 \rightarrow \text { odd } \\ & R=\{x-y=\text { odd }+ \text { veor } x-y=2\} \quad 0,4,6,8,10 \rightarrow \text { even } \\ & { }^3 C_1 \cdot{ }^5 C_1=15+(6,4),(8,6),(10,8),(9,7) \end{aligned}

\operatorname{Min}^{\mathrm{m}}   ordered pairs to be added must be

: 15+4=19

Posted by

Suraj Bhandari

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