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 $R_1$ and $R_2$, respectively, in order to make the relations symmetric, then $M+N$ equals

Accepted Answer

$$\begin{aligned}
& R_1=\{(x, y): 2 x-3 y=2\} \
& R_2=\{(x, y):-5 x+4 y=0\} \
& 2 x-3 y=2
\end{aligned}$$

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

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

For $R_2-5 x+4 y=0$

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

$$R_2=\{(4,5),(8,10),(12,15),(16,20)\}$$

For symmetric we need to add 4 element as

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

Solutions · 1