The relation R is defined on the set N by R = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the relation $R = \{(x, y) | x, y \in N, 2x + y = 41\}$.For $R$ to be reflexive,$(x, x) \in R$ for all $x \in N$. This implies $2x + x = 41$

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Given the relation $R = \{(x, y) | x, y \in N, 2x + y = 41\}$.For $R$ to be reflexive,$(x, x) \in R$ for all $x \in N$. This implies $2x + x = 41$,so $3x = 41$,which gives $x = 41/3 
otin N$. Thus,$R$ is not reflexive.For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x) \in R$. Let $(1, 39) \in R$ because $2(1) + 39 = 41$. However,$(39, 1) 
otin R$ because $2(39) + 1 = 79 
eq 41$. Thus,$R$ is not symmetric.For $R$ to be transitive,if $(x, y) \in R$ and $(y, z) \in R$,then $(x, z) \in R$. Let $(1, 39) \in R$ and $(39, z) \in R$. For $(39, z) \in R$,$2(39) + z = 41$,which gives $z = 41 - 78 = -37 
otin N$. Since there is no such $z \in N$,the condition for transitivity fails. Thus,$R$ is not transitive.Therefore,$R$ is none of these.

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

Similar Questions

The relation "less than" in the set of natural numbers is

On the set of all real numbers,a relation $R$ is defined as $a \, R \, b$ if and only if $|a - b| \le 1$. Then $R$ is:

$A$ relation $R$ on the set of natural numbers is defined as $\{(a, b) : |a - b| = 3\}$. Then $R$ is:

Let the relation $R_{1}$ be defined on $R$ as $a R_{1} b$ if $1+ab > 0$. Then

Let $A = \{2, 3, 4\}$ and $B = \{8, 9, 12\}$. Then the number of elements in the relation $R = \{((a_1, b_1), (a_2, b_2)) \in (A \times B) \times (A \times B) : a_1 \text{ divides } b_2 \text{ and } a_2 \text{ divides } b_1\}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module