Let R be a relation on N times N defined by ( | vedclass

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

Question

(D) The relation is defined as $(a, b) R (c, d) \iff ad(b-c) = bc(a-d)$.$1$. Reflexive: For $(a, b) R (a, b)$,we need $ab(b-a) = ba(a-b)$. This simpli

Vedclass · Accepted Answer

symmetric and transitive but not reflexive

Vedclass · Answer

(D) The relation is defined as $(a, b) R (c, d) \iff ad(b-c) = bc(a-d)$.$1$. Reflexive: For $(a, b) R (a, b)$,we need $ab(b-a) = ba(a-b)$. This simplifies to $ab(b-a) = -ab(b-a)$,which is only true if $ab(b-a) = 0$. Since $a, b \in N$,this is not true for all $(a, b)$. Thus,$R$ is not reflexive.$2$. Symmetric: If $(a, b) R (c, d)$,then $ad(b-c) = bc(a-d) \Rightarrow adb - adc = bca - bcd \Rightarrow adb + bcd = bca + adc$. Dividing by $abcd$,we get $\frac{1}{c} - \frac{1}{d} = \frac{1}{a} - \frac{1}{b}$. This is symmetric because swapping $(a, b)$ and $(c, d)$ yields the same condition. Thus,$R$ is symmetric.$3$. Transitive: The condition $\frac{1}{b} - \frac{1}{a} = \frac{1}{d} - \frac{1}{c}$ implies that the relation is an equivalence relation (if defined on $N 	imes N$ such that $a, b 
eq 0$). However,checking the specific form $ad(b-c) = bc(a-d)$,it is equivalent to $\frac{b-c}{bc} = \frac{a-d}{ad} \Rightarrow \frac{1}{c} - \frac{1}{b} = \frac{1}{d} - \frac{1}{a}$. This is transitive. Since it is symmetric and transitive but fails reflexivity (e.g.,$(1, 1) R (1, 1)$ holds,but for general $(a, b)$,it depends on the domain),the correct classification is symmetric and transitive.

Let $R$ be a relation on $N \times N$ defined by $(a, b) R (c, d)$ if and only if $ad(b-c) = bc(a-d)$. Then $R$ is

Similar Questions

For a set $A = \{1, 2, 3\}$,a relation $R = \{(1, 2), (2, 3)\}$ is defined. What is the minimum number of ordered pairs that must be added to $R$ to make it an equivalence relation?

If $A = \{x \in Z^+ : x < 10\}$ and $x$ is a multiple of $3$ or $4$,where $Z^+$ is the set of positive integers,then the total number of symmetric relations on $A$ is

Let $A = \{2, 3, 4\}$ and $B = \{8, 9, 12\}$. Then the number of elements in the relation $R = \{((a_1, b_1), (a_2, b_2)) \in (A \times B) \times (A \times B) : a_1 \text{ divides } b_2 \text{ and } a_2 \text{ divides } b_1\}$ is:

The number of relations,defined on the set ${a, b, c, d}$,which are both reflexive and symmetric,is equal to:

Consider the relations $R_1$ and $R_2$ defined as $a R_1 b \Leftrightarrow a^2+b^2=1$ for all $a, b \in R$ and $(a, b) R_2 (c, d) \Leftrightarrow a+d=b+c$ for all $(a, b), (c, d) \in N \times N$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module