Let A=-2,-1,0,1,2,3,4.

Question

Let $A={-2,-1,0,1,2,3,4}$. Let R be a relation on A defined by $xRy$ if and only if $2x+y\leqslant 2$. 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

For each $x\in A$, find $y\in A$ with $y \leq 2-2x$:

- $x=-2$: $y\leq6$ 	o  all 7 elements of $A$
- $x=-1$: $y\leq4$ 	o  all 7 elements of $A$
- $x=0$: $y\leq2$ 	o  $\{-2,-1,0,1,2\}$, 5 values
- $x=1$: $y\leq0$ 	o  $\{-2,-1,0\}$, 3 values
- $x=2$: $y\leq-2$ 	o  $\{-2\}$, 1 value
- $x=3$: $y\leq-4$ 	o  none
- $x=4$: $y\leq-6$ 	o  none

$l = 7+7+5+3+1 = 23$

Reflexive: need $(x,x)$, i.e. $3x\leq2 \Rightarrow x\leq	frac23$. Already true for $x=-2,-1,0$. Missing for $x=1,2,3,4$: $m=4$.

Symmetric: Checking each pair $(x,y)$ in $R$ against its reverse (using the listed pairs above), the pairs whose reverse is *not* already in $R$ are $(3,-2), (4,-2), (2,-1), (3,-1), (4,-1), (2,0)$ --- six pairs, so $n=6$.

$l+m+n = 23+4+6 = 33$

Solutions · 1