The relation R = {(x, y) : x, y in mathbb{Z} | vedclass

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

Question

(C) Given relation $R = \{(x, y) : x, y \in \mathbb{Z} 	ext{ and } x + y 	ext{ is even} \}$.$1$. Reflexivity: For any $x \in \mathbb{Z}$,$x + x = 2x

Vedclass · Accepted Answer

an equivalence relation

Vedclass · Answer

(C) Given relation $R = \{(x, y) : x, y \in \mathbb{Z} 	ext{ and } x + y 	ext{ is even} \}$.$1$. Reflexivity: For any $x \in \mathbb{Z}$,$x + x = 2x$,which is always an even integer. Thus,$(x, x) \in R$ for all $x \in \mathbb{Z}$. So,$R$ is reflexive.$2$. Symmetry: If $(x, y) \in R$,then $x + y$ is even. Since $x + y = y + x$,$y + x$ is also even. Thus,$(y, x) \in R$. So,$R$ is symmetric.$3$. Transitivity: If $(x, y) \in R$ and $(y, z) \in R$,then $x + y$ is even and $y + z$ is even. The sum of two even numbers is even,so $(x + y) + (y + z) = x + z + 2y$ is even. Since $2y$ is even,$x + z$ must be even. Thus,$(x, z) \in R$. So,$R$ is transitive.Since the relation is reflexive,symmetric,and transitive,it is an equivalence relation.

The relation $R = \{(x, y) : x, y \in \mathbb{Z} \text{ and } x + y \text{ is even} \}$ is :

Similar Questions

Show that the relation $R$ in the set $Z$ of integers given by $R = \{(a, b) : 2 \text{ divides } a - b\}$ is an equivalence relation.

Let $A = \{2, 4, 6, 8\}$. $A$ relation $R$ on $A$ is defined by $R = \{(2, 4), (4, 2), (4, 6), (6, 4)\}$. Then $R$ is:

Among the relations $S = \{(a, b) : a, b \in R - \{0\}, 2 + \frac{a}{b} > 0\}$ and $T = \{(a, b) : a, b \in R, a^2 - b^2 \in Z\}$,which of the following is true?

The maximum number of equivalence relations on the set $A = \{1, 2, 3, 4\}$ is $N$. Then -

Let $A$ be the set of all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ and $R$ be a relation on $A$ such that $R =\{( f , g ): f(0)= g (1) \text{ and } f(1)= g (0)\}$. Then $R$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module