Let R_{1} = {(a, b) in N times N : |a - b| le | vedclass

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

Question

(B) For $R_{1} = \{(a, b) \in N 	imes N : |a - b| \leq 13\}$:$(i)$ Reflexive: $|a - a| = 0 \leq 13$,so $(a, a) \in R_{1}$. It is reflexive.(ii) Symme

Vedclass · Accepted Answer

Neither $R_{1}$ nor $R_{2}$ is an equivalence relation.

Vedclass · Answer

(B) For $R_{1} = \{(a, b) \in N 	imes N : |a - b| \leq 13\}$:$(i)$ Reflexive: $|a - a| = 0 \leq 13$,so $(a, a) \in R_{1}$. It is reflexive.(ii) Symmetric: If $|a - b| \leq 13$,then $|b - a| = |-(a - b)| = |a - b| \leq 13$. So $(b, a) \in R_{1}$. It is symmetric.(iii) Transitive: Let $(1, 10) \in R_{1}$ and $(10, 20) \in R_{1}$. Here $|1 - 10| = 9 \leq 13$ and $|10 - 20| = 10 \leq 13$. However,$|1 - 20| = 19 
ot\leq 13$. Thus,$(1, 20) 
otin R_{1}$. $R_{1}$ is not transitive,hence not an equivalence relation.For $R_{2} = \{(a, b) \in N 	imes N : |a - b| 
eq 13\}$:$(i)$ Reflexive: $|a - a| = 0 
eq 13$. So $(a, a) \in R_{2}$. It is reflexive.(ii) Symmetric: If $|a - b| 
eq 13$,then $|b - a| = |a - b| 
eq 13$. So $(b, a) \in R_{2}$. It is symmetric.(iii) Transitive: Let $(1, 14) \in R_{2}$ and $(14, 1) \in R_{2}$. Here $|1 - 14| = 13$ and $|14 - 1| = 13$. Wait,$(1, 14) 
otin R_{2}$ and $(14, 1) 
otin R_{2}$. Let's test $(1, 2) \in R_{2}$ and $(2, 15) \in R_{2}$. $|1 - 2| = 1 
eq 13$ and $|2 - 15| = 13$. Since $(2, 15) 
otin R_{2}$,this is not a valid counterexample. Consider $(1, 2) \in R_{2}$ and $(2, 14) \in R_{2}$. $|1 - 2| = 1 
eq 13$ and $|2 - 14| = 12 
eq 13$. But $|1 - 14| = 13$,so $(1, 14) 
otin R_{2}$. Thus,$R_{2}$ is not transitive,hence not an equivalence relation.Therefore,neither $R_{1}$ nor $R_{2}$ is an equivalence relation.

Let $R_{1} = \{(a, b) \in N \times N : |a - b| \leq 13\}$ and $R_{2} = \{(a, b) \in N \times N : |a - b| \neq 13\}$. Then on $N$:

Similar Questions

Let $A = \{1, 2, 3, \ldots, 20\}$. Let $R_1$ and $R_2$ be two relations on $A$ such that $R_1 = \{(a, b) : b \text{ is divisible by } a\}$ and $R_2 = \{(a, b) : a \text{ is an integral multiple of } b\}$. Then,the number of elements in $R_1 - R_2$ is equal to . . . . . . .

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?
$(a, b) \in R, (b, c) \in R$ implies $(a, c) \in R$

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

Let $R$ be a relation on the set of all natural numbers $\mathbb{N}$ defined by $aRb \iff a \text{ divides } b^2$. Which of the following properties does $R$ satisfy?
$I.$ Reflexivity
$II.$ Symmetry
$III.$ Transitivity

Let $A = \{1, 2, 3, 4, 5\}$. $A$ relation $R$ on $A$ is defined by $R = \{(x, y) | x, y \in A \text{ and } x < y\}$. Then $R$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module