Let A = {-3, -2, -1, 0, 1, 2, 3} and R be a r | vedclass

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

Question

(B) The relation $R$ is defined on $A = \{-3, -2, -1, 0, 1, 2, 3\}$ by $2x - y \in \{0, 1\}$.Case $1$: $2x - y = 0 \implies y = 2x$.For $x = -1, y = -

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(B) The relation $R$ is defined on $A = \{-3, -2, -1, 0, 1, 2, 3\}$ by $2x - y \in \{0, 1\}$.Case $1$: $2x - y = 0 \implies y = 2x$.For $x = -1, y = -2$; for $x = 0, y = 0$; for $x = 1, y = 2$.Pairs: $(-1, -2), (0, 0), (1, 2)$.Case $2$: $2x - y = 1 \implies y = 2x - 1$.For $x = -1, y = -3$; for $x = 0, y = -1$; for $x = 1, y = 1$; for $x = 2, y = 3$.Pairs: $(-1, -3), (0, -1), (1, 1), (2, 3)$.Thus,$R = \{(-1, -2), (0, 0), (1, 2), (-1, -3), (0, -1), (1, 1), (2, 3)\}$.The number of elements $l = 7$.For $R$ to be reflexive,we need $(x, x) \in R$ for all $x \in A$. Currently,only $(0, 0)$ and $(1, 1)$ are present. We need to add $(-3, -3), (-2, -2), (-1, -1), (2, 2), (3, 3)$. So,$m = 5$.For $R$ to be symmetric,if $(x, y) \in R$,then $(y, x)$ must be in $R$. The elements are $(-1, -2), (1, 2), (-1, -3), (0, -1), (2, 3)$. Their inverses are $(-2, -1), (2, 1), (-3, -1), (-1, 0), (3, 2)$. None of these are in $R$. Thus,we need to add $5$ elements. So,$n = 5$.Therefore,$l + m + n = 7 + 5 + 5 = 17$.

Similar Questions

Let $N$ be the set of natural numbers greater than $100$. Define the relation $R$ on $N$ by: $R = \{(x, y) \in N \times N : \text{the numbers } x \text{ and } y \text{ have at least two common divisors}\}.$ Then $R$ is-

Show that the relation $R$ in the set $A = \{x \in Z : 0 \leq x \leq 12\},$ given by $R = \{(a, b) : |a - b| \text{ is a multiple of } 4\},$ is an equivalence relation. Find the set of all elements related to $1$.

Consider the following two binary relations on the set $A = \{a, b, c\}$: $R_1 = \{(c, a), (b, b), (a, c), (c, c), (b, c), (a, a)\}$ and $R_2 = \{(a, b), (b, a), (c, c), (c, a), (a, a), (b, b), (a, c)\}$. Then

Check whether the relation $R$ defined in the set $\{1, 2, 3, 4, 5, 6\}$ as $R = \{(a, b) : b = a + 1\}$ is reflexive,symmetric,or transitive.

Let $R = \{(x, y) \in N \times N : \log_e(x + y) \leq 2\}$. Then the minimum number of elements, required to be added in $R$ to make it a transitive relation, is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module