Let A=2,3,5,7,9.

Question

Let $A={2,3,5,7,9}$. Let $R$ be the relation on $A$ defined by $xRy$ if and only if $2x\le 3y$. Let $l$ be the number of elements in $R$, and $m$ be the minimum number of elements required to be added in $R$ to make it a symmetric relation. Then $l+m$ is equal to:

Accepted Answer

For each $x$, find $y \in A$ with $2x \leq 3y$:

- $x=2$: $4\leq 3y \Rightarrow y\in\{2,3,5,7,9\}$ (5 pairs)
- $x=3$: $6\leq 3y \Rightarrow y\in\{2,3,5,7,9\}$ (5 pairs)
- $x=5$: $10\leq 3y \Rightarrow y\in\{5,7,9\}$ (3 pairs)
- $x=7$: $14\leq 3y \Rightarrow y\in\{5,7,9\}$ (3 pairs)
- $x=9$: $18\leq 3y \Rightarrow y\in\{7,9\}$ (2 pairs)

$l = 5+5+3+3+2 = 18$

Symmetry check --- for each $(x,y)\in R$, $x
eq y$, check whether $(y,x)\in R$:

- $(2,3)$: need $(3,2)$: $6\leq6$  already present
- $(2,5)$: need $(5,2)$: $10\leq6$  --- add
- $(2,7)$: need $(7,2)$: $14\leq6$  --- add
- $(2,9)$: need $(9,2)$: $18\leq6$  --- add
- $(3,5)$: need $(5,3)$: $10\leq9$  --- add
- $(3,7)$: need $(7,3)$: $14\leq9$  --- add
- $(3,9)$: need $(9,3)$: $18\leq9$  --- add
- $(5,7)$: need $(7,5)$: $14\leq15$  already present
- $(5,9)$: need $(9,5)$: $18\leq15$  --- add
- $(7,9)$: need $(9,7)$: $18\leq21$  already present

Pairs to add: 7, so $m=7$.

$l+m = 18+7 = 25$

Solutions · 1