Let A = {0, 3, 4, 6, 7, 8, 9, 10} and R be th | vedclass

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

Question

(B) The set $A$ is given as $A = \{0, 3, 4, 6, 7, 8, 9, 10\}$. The number of odd elements is $3$ $(\{3, 7, 9\})$ and the number of even elements is $5

Vedclass · Accepted Answer

$19$

Vedclass · Answer

(B) The set $A$ is given as $A = \{0, 3, 4, 6, 7, 8, 9, 10\}$. The number of odd elements is $3$ $(\{3, 7, 9\})$ and the number of even elements is $5$ $(\{0, 4, 6, 8, 10\})$.The relation $R$ contains pairs $(x, y)$ such that $x - y$ is an odd positive integer or $x - y = 2$.$1$. Pairs where $x - y$ is an odd positive integer: Since $x - y$ is odd,one must be odd and the other even. There are $3 	imes 5 = 15$ such pairs where $x > y$.$2$. Pairs where $x - y = 2$: These are $(6, 4), (8, 6), (10, 8), (9, 7)$. There are $4$ such pairs where $x > y$.Total pairs in $R$ with $x > y$ is $15 + 4 = 19$.For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x)$ must also be in $R$. Since all $19$ pairs currently satisfy $x > y$,their symmetric counterparts $(y, x)$ where $y Therefore,we must add $19$ elements to $R$ to make it symmetric.

Similar Questions

If $R$ and $R^1$ are equivalence relations on a set $A$, then which of the following is also an equivalence relation?

Let $R_{1}$ and $R_{2}$ be two relations defined as follows:
$R_{1} = \{(a, b) \in \mathbb{R}^{2} : a^{2} + b^{2} \in \mathbb{Q}\}$ and $R_{2} = \{(a, b) \in \mathbb{R}^{2} : a^{2} + b^{2} \notin \mathbb{Q}\}$
where $\mathbb{Q}$ is the set of all rational numbers. Then:

Let $\rho_{1}$ and $\rho_{2}$ be two equivalence relations defined on a non-void set $S$. Then

Let $R = \{(1, 3), (2, 2), (3, 2)\}$ and $S = \{(2, 1), (3, 2), (2, 3)\}$ be two relations on the set $A = \{1, 2, 3\}$. Find $R \circ S^{-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module