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

Question

Let $A={-3,-2,-1,0,1,2,3}$ and R be a relation on A defined by $xRy$ if and only if $2x-y\in {0,1}$. 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 $l+m+n$ is equal to:

Accepted Answer

$xRy \iff y=2x$ or $y=2x-1$. Checking $x\in A$ against $A$'s range $[-3,3]$:

- $x=-1$: $y=-2$ or $y=-3$ 	o  $(-1,-2),(-1,-3)$
- $x=0$: $y=0$ or $y=-1$ 	o  $(0,0),(0,-1)$
- $x=1$: $y=2$ or $y=1$ 	o  $(1,2),(1,1)$
- $x=2$: $y=4$ (invalid) or $y=3$ 	o  $(2,3)$
- $x=-3,-2,3$: no valid $y$ in $A$

$R = \{(-1,-2),(-1,-3),(0,0),(0,-1),(1,2),(1,1),(2,3)\}$, so $l=7$.

Reflexive: need $(x,x)$, i.e. $x\in\{0,1\}$ (both already present). Missing: $x=-3,-2,-1,2,3$ 	o  $m=5$.

Symmetric: for each non-self pair, check the reverse --- $(-1,-2)	o(-2,-1)$, $(-1,-3)	o(-3,-1)$, $(0,-1)	o(-1,0)$, $(1,2)	o(2,1)$, $(2,3)	o(3,2)$ --- all missing, giving $n=5$.

$l+m+n = 7+5+5 = 17$

Solutions · 1