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Let $A=\{1,2,3,4,5\}$. Let $\mathrm{R}$ be a relation on $\mathrm{A}$ defined by $x \mathrm{R} y$ if and only if $4 x \leq 5 \mathrm{y}$. Let $\mathrm{m}$ be the number of elements in $\mathrm{R}$ and $\mathrm{n}$ be the minimum number of elements from $\mathrm{A} \times \mathrm{A}$ that are required to be added to $\mathrm{R}$ to make it a symmetric relation. Then $\mathrm{m}+\mathrm{n}$ is equal to :
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2192 Upvotes
Verified Answer
The correct answer is:
25
Given : $4 x \leq 5 y$
then
$\begin{aligned}
\mathrm{R}=\{ & (1,1),(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4) \\
& (2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)\}
\end{aligned}$
i.e. 16 elements.
i.e. $\mathrm{m}=16$
Now to make R a symmetric relation add
$\{(2,1)(3,2)(4,3)(3,1)(4,2)(5,3)(4,1)(5,2)(5,1)\}$
i.e. $n=9$
So $m+n=25$
then
$\begin{aligned}
\mathrm{R}=\{ & (1,1),(1,2),(1,3),(1,4),(1,5),(2,2),(2,3),(2,4) \\
& (2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)\}
\end{aligned}$
i.e. 16 elements.
i.e. $\mathrm{m}=16$
Now to make R a symmetric relation add
$\{(2,1)(3,2)(4,3)(3,1)(4,2)(5,3)(4,1)(5,2)(5,1)\}$
i.e. $n=9$
So $m+n=25$
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