Let P = { (x, y) | x^2 + y^2 = 1, x, y in mat | vedclass

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

Question

(B) relation $P$ on a set $A$ is reflexive if $(x, x) \in P$ for all $x \in A$. Here,$x^2 + x^2 = 1 \Rightarrow 2x^2 = 1 \Rightarrow x = \pm \frac{1}{

Vedclass · Accepted Answer

Symmetric

Vedclass · Answer

(B) relation $P$ on a set $A$ is reflexive if $(x, x) \in P$ for all $x \in A$. Here,$x^2 + x^2 = 1 \Rightarrow 2x^2 = 1 \Rightarrow x = \pm \frac{1}{\sqrt{2}}$. Since this is not true for all $x \in \mathbb{R}$,$P$ is not reflexive.$A$ relation $P$ is symmetric if $(x, y) \in P \Rightarrow (y, x) \in P$. Given $x^2 + y^2 = 1$,it follows that $y^2 + x^2 = 1$. Thus,if $(x, y) \in P$,then $(y, x) \in P$. Hence,$P$ is symmetric.$A$ relation $P$ is transitive if $(x, y) \in P$ and $(y, z) \in P \Rightarrow (x, z) \in P$. Let $(x, y) = (1, 0)$ and $(y, z) = (0, 1)$. Both are in $P$ because $1^2 + 0^2 = 1$ and $0^2 + 1^2 = 1$. However,$(x, z) = (1, 1)$ is not in $P$ because $1^2 + 1^2 = 2 
eq 1$. Thus,$P$ is not transitive.

Let $P = \{ (x, y) | x^2 + y^2 = 1, x, y \in \mathbb{R} \}$. Then $P$ is:

Similar Questions

$A$ relation $R$ defined on a set $A$ is anti-symmetric if $(a, b) \in R$ and $(b, a) \in R$ implies:

Let $R$ be a relation from $A = \{2, 3, 4, 5\}$ to $B = \{3, 6, 7, 10\}$ defined by $R = \{(a, b) \mid a \text{ divides } b, a \in A, b \in B\}$. Then,the number of elements in $R^{-1}$ will be:

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

Let $T$ be the set of all triangles in a plane with $R$ a relation in $T$ given by $R = \{(T_1, T_2) : T_1 \text{ is congruent to } T_2\}$. Show that $R$ is an equivalence relation.

The relation $R = \{(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)\}$ on set $A = \{1, 2, 3\}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module