Let A=-3,-2,-1,0,1,2,3 and R be a relation on A defined by x R y if and only if 2 x-y 0,1.

Question

Let $\mathrm{A}=\{-3,-2,-1,0,1,2,3\}$ and R be a relation on A defined by $x \mathrm{R} y$ if and only if $2 x-y \in\{0,1\}$. Let $l$ be the number of elements in $R$. Let $m$ and $n$ be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then $l+\mathrm{m}+\mathrm{n}$ is equal to:

Accepted Answer

$$\begin{aligned}
&\begin{aligned}
& x R y \Leftrightarrow 2 x-y \in\{0,1\} \
& \Rightarrow \quad y=2 x \text { or } y=2 x-1 \
& A=\{-3,-2,-1,0,1,2,3\} \
& \mathrm{R}=\{(-1,-2),(0,0),(1,2),(-1,-3),(0,-1),(1,1), \
& (2,3)\} \
& \Rightarrow \quad I=7
\end{aligned}\
\&\text { For } R \text { to be reflexive }(0,0),(1,1) \in R
\end{aligned}$$

But other $(a, a)$ such that $2 a-a \in\{0,1\}$

$$\Rightarrow \quad a \in\{0,1\}$$

5 other pairs needs to be added $\Rightarrow m=5$

$x R y \Rightarrow y R x$ to be symmetric

$(-1,-2) \Rightarrow(-2,-1)$

$(1,2) \Rightarrow(2,1)$

$(-1,-3) \Rightarrow(-3,-1)$

$(0,-1) \Rightarrow(-1,0)$

$(2,3) \Rightarrow(3,2) \Rightarrow 5$ needs to be added, $n=5$

$\Rightarrow \quad l+m+n=17$

Solutions · 1