On the set R of real numbers, we define x P y | vedclass

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

Question

(D) For every real number $x$, $x^2 \geq 0$. $	herefore (x, x) \in P$. Hence, $P$ is reflexive. Now, let $(x, y) \in P$. $\Rightarrow xy \geq 0$. $\R

Vedclass · Accepted Answer

reflexive and symmetric but not transitive

Vedclass · Answer

(D) For every real number $x$, $x^2 \geq 0$. $	herefore (x, x) \in P$. Hence, $P$ is reflexive. Now, let $(x, y) \in P$. $\Rightarrow xy \geq 0$. $\Rightarrow yx \geq 0$. $	herefore (y, x) \in P$. Hence, $P$ is symmetric. Again, consider $(-1, 0) \in P$ because $(-1)(0) = 0 \geq 0$, and $(0, 2) \in P$ because $(0)(2) = 0 \geq 0$. However, $(-1, 2) 
otin P$ because $(-1)(2) = -2 Therefore, $P$ is not transitive. Thus, the relation $P$ is reflexive and symmetric but not transitive.

On the set $R$ of real numbers, we define $x P y$ if and only if $x y \geq 0$. Then, the relation $P$ is

Similar Questions

Let $n$ be a fixed positive integer. Define a relation $R$ on the set $Z$ of integers by $aRb \Leftrightarrow n | (a - b)$. Then $R$ is

Let a set $A = A_{1} \cup A_{2} \cup \ldots \cup A_{k}$,where $A_{i} \cap A_{j} = \phi$ for $i \neq j$ and $1 \leq i, j \leq k$. Define the relation $R$ from $A$ to $A$ by $R = \{(x, y) : y \in A_{i} \text{ if and only if } x \in A_{i}, 1 \leq i \leq k\}$. Then,$R$ is

Let $R$ be a relation from $N$ to $N$ defined by $R = \{(a, b) : a, b \in N \text{ and } a = b^2\}$. Is the following statement true?
$(a, a) \in R$,for all $a \in N$

Let $R$ be the relation in the set $\{1, 2, 3\}$ given by $R = \{(1, 1), (2, 2), (3, 3)\}$. Choose the correct answer.

Determine whether the following relation is reflexive,symmetric,and transitive:
Relation $R$ in the set $A$ of human beings in a town at a particular time given by $R = \{(x, y): x \text{ is the father of } y\}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module