Question

Let $R$ a relation on the set $N$ be defined by $\{(x,y)|x,y\in N,2x+y=41\}$. Then $R$ is

• A. Reflexive
• B. Symmetric
• C. Transitive
• D. None of these

Answer 1

The correct option is D None of these

Given R:{(x,y):x∈Ny∈N,2x+y=41}R:{(x,y):x∈Ny∈N,2x+y=41}

2x+y=412x+y=41

y=41−2xy=41−2x

Since y∈Ny>0y∈Ny>0

41−2x>041−2x>0

−2x>−41−2x>−41

2x<412x<41

x<412x<412

Since x∈N.x∈N. x can take values ={1,2,3....20}={1,2,3....20}

Domain of R={1,2,3.....20}R={1,2,3.....20}

Range of R is y=41−2xforx={1,2,3....20}y=41−2xforx={1,2,3....20}

Range of R={1,3,.......,37,39}R={1,3,.......,37,39}

When x=1y=41−2=39x=1y=41−2=39

When x=2y=41−2x×2=39x=2y=41−2x×2=39

When x=20y=40−2×20=1x=20y=40−2×20=1

Domain of f={1,2.....20}f={1,2.....20}

Range of f={1,3......37,39}f={1,3......37,39}

R is not reflexive

Since (1,1) does not satisfy

2x+y=412x+y=41

2×1+1≠412×1+1≠41

R is not symmetric since (1,39)∈R(1,39)∈R satisfies 2x+y=41but(39,1)∉R2x+y=41but(39,1)∉R

Since 2×39+1≠412×39+1≠41

but 2×1+39=412×1+39=41

Solution:R is not transitive since (1,39) ∈∈ but no values for x=39. satisfies the given relation