Let A=1,2,3,4,5.

Question

Let $A={1,2,3,4,5}$. Let $R$ be a relation on $A$ defined by $xRy$ if and only if $4x\le 5y$. Let $m$ be the number of elements in $R$ and $n$ be the minimum number of elements from $A	imes A$ that are required to be added to R to make it a symmetric relation. Then m + n is equal to :

Accepted Answer

$\begin{gathered} A={1,2,3,4,5} \ xRy\Leftrightarrow 4x\le 5y \ 4x\le 5y\, \Rightarrow \, \frac{x}{y}\le \frac{5}{4}\, \Rightarrow \frac{x}{y}\le 1.25 \end{gathered}$

$\begin{gathered} R={(1,2),(1,3),(1,4),(1,5),(1,1),(2,2),(2,3),(2,4), \ (2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)} \ 	herefore \, n(R)=m=16 \end{gathered}$

Elements to be added to $R$ to make it symmetric

$(1,2)\in R\, \Rightarrow \, (2,1)$ should be added

Similarly, $(3,1),(4,1),(5,1),(3,2),(4,2),(5,2),(4,3),(5,3)$

$	herefore 9$ elements should be added

$\begin{gathered} 	herefore n=9 \ 	herefore m+n=25 \end{gathered}$

Solutions · 1