Let the relations R_1 and R_2 on the set [X=1,2,3, , 20] be given by [R_1=(x, y): 2 x-3 y=…

Question

Let the relations $R_1$ and $R_2$ on the set $$X=\{1,2,3, \ldots, 20\}$$ be given by $$R_1=\{(x, y): 2 x-3 y=2\}$$ and $$R_2=\{(x, y):-5 x+4 y=0\}$$. If $M$ and $N$ be the minimum number of elements required to be added in $R_1$ a...

Accepted Answer

$\begin{gathered} {R}_{1}={(x,y):2x-3y=2} \ {R}_{2}={(x,y):-5x+4y=0} \ 2x-3y=2 \end{gathered}$

So $2x$ and $3y$ both has to be even or odd simultaneously and $2x$ can't be odd so $2x$ and $3y$ both will be even

${R}_{1}={(4,2),(7,4),(10,6),(13,8),(16,10),(19,12)}$

For symmetric we need to add 6 elements as

$\begin{gathered} (2,4),(4,7),(6,10),(8,13),(10,16),(12,19) \ M=6 \end{gathered}$

For ${R}_{2}-5x+4y=0$

$5x$ and $4y$ has to be equal $4y$ is always even so $5x$ will also be even

${R}_{2}={(4,5),(8,10),(12,15),(16,20)}$

For symmetric we need to add 4 element as

$\begin{gathered} (5,4)(10,8)(15,12)(20,16) \ N=4 \ M+N=6+4=10 \end{gathered}$

Solutions · 1