Let the relation R on the set M=1,2,3, ,16 be given by R=(x,y):4y=5x-3,x,y M.

Question

Let the relation R on the set $M={1,2,3,\ldots ,16}$ be given by $R={(x,y):4y=5x-3,x,y\in M}$.

Then the minimum number of elements required to be added in R , in order to make the relation symmetric, is equal to

Accepted Answer

$4y = 5x-3 \Rightarrow y = \dfrac{5x-3}{4}$. For $y$ to be an integer, $5x\equiv3\pmod4$, i.e. $x\equiv3\pmod4$ (since $5\equiv1\pmod4$). So $x \in \{3,7,11,15\}$ within $1$--$16$.

- $x=3\Rightarrow y=3$: pair $(3,3)$
- $x=7\Rightarrow y=8$: pair $(7,8)$
- $x=11\Rightarrow y=13$: pair $(11,13)$
- $x=15\Rightarrow y=18$: rejected, $18
otin M$

So $R = \{(3,3),(7,8),(11,13)\}$.

$(3,3)$ is already symmetric (it's a self-pair). For $(7,8)$ we need to add $(8,7)$, and for $(11,13)$ we need to add $(13,11)$.

Minimum elements to add: 2.

Solutions · 1