Let N denote the set of all natural numbers. Define two binary relations on N as R1 = (x, y) ϵ N × N: 2x + y = 10 and R2 = (x, y) ϵ N × N: x + 2y = 10 . Then :

Question

Let N denote the set of all natural numbers. Define two binary relations on N as R1 = (x, y) ϵ N × N: 2x + y = 10 and R2 = (x, y) ϵ N × N: x + 2y = 10 . Then : (A) Range of R1 is 2, 4, 8  (B) Range of R2 is 1, 2, 3, 4  (C) Both R1 and R2 are symmetric relations (D) Both R1 and R2 are transitive relations

Tardigrade Admin · Accepted Answer

Given: N is set of all natural numbers R1= (x, y) ∈ N × N: 2 x+y=10 and R2= (x, y) ∈ N × N: x+2 y=10 R1= (1,8),(2,6),(3,4),(4,2) and R2= (8,1),(6,2),(4,3),(2,4) Therefore, range of R2= 1,2,3,4

Q. Let N denote the set of all natural numbers. Define two binary relations on N as $R_{1} = {(x, y) ϵ N \times N : 2 x + y = 10}$ and $R_{2} = {(x, y) ϵ N \times N : x + 2 y = 10}$ . Then :

Range of $R_{1}$ is ${2, 4, 8}$

Range of $R_{2}$ is ${1, 2, 3, 4}$

Both $R_{1}$ and $R_{2}$ are symmetric relations

Both $R_{1}$ and $R_{2}$ are transitive relations

Given: $N$ is set of all natural numbers
$R_{1} = {(x, y) \in N \times N : 2 x + y = 10}$ and $R_{2} = {(x, y) \in N \times N : x + 2 y = 10}$
$R_{1} = {(1, 8), (2, 6), (3, 4), (4, 2)}$ and $R_{2} = {(8, 1), (6, 2), (4, 3), (2, 4)}$
Therefore, range of $R_{2} = {1, 2, 3, 4}$

Q. Let N denote the set of all natural numbers. Define two binary relations on N as R1​={(x,y)ϵN×N:2x+y=10} and R2​={(x,y)ϵN×N:x+2y=10}. Then :

Range of R1​ is {2,4,8}

Range of R2​ is {1,2,3,4}

Both R1​ and R2​ are symmetric relations

Both R1​ and R2​ are transitive relations