(B) Since $R$ is an equivalence relation on set $A$,it must be reflexive. By the definition of reflexivity,$(a, a) \in R$ for all $a \in A$. Since the

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(B) Since $R$ is an equivalence relation on set $A$,it must be reflexive. By the definition of reflexivity,$(a, a) \in R$ for all $a \in A$. Since there are $n$ elements in set $A$,there must be at least $n$ ordered pairs of the form $(a, a)$ in $R$. Therefore,the number of ordered pairs in $R$ is greater than or equal to $n$.

Class 12 Mathematics Relation and Function Types of Relations

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Let $R$ be an equivalence relation on a finite set $A$ having $n$ elements. Then the number of ordered pairs in $R$ is:

A
Less than $n$
B
Greater than or equal to $n$
C
Equal to or less than $n$
D
None of these

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Relation and Function

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Types of Relations

Let $N$ denote the set of all natural numbers. Define two binary relations on $N$ as $R_1 = \{(x,y) \in N \times N : 2x + y = 10\}$ and $R_2 = \{(x,y) \in N \times N : x + 2y = 10\}$. Then

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In the set $N$,a relation $R$ is defined as $aRb \Leftrightarrow b$ is divisible by $a$. Then $R$ is:

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Let $I$ be the set of positive integers. $R$ is a relation on the set $I$ given by $R = \{(a, b) \in I \times I \mid \log_2(a/b) \text{ is a non-negative integer} \}$. Then $R$ is:

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Let $S$ be the set of all real numbers. Then the relation $R = \{(a, b) : 1 + ab > 0\}$ on $S$ is

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Let $R = \{(1, 3), (2, 2), (3, 2)\}$ and $S = \{(2, 1), (3, 2), (2, 3)\}$ be two relations on the set $A = \{1, 2, 3\}$. Find $R \circ S^{-1}$.

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Let $R$ be an equivalence relation on a finite set $A$ having $n$ elements. Then the number of ordered pairs in $R$ is:

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Let $N$ denote the set of all natural numbers. Define two binary relations on $N$ as $R_1 = \{(x,y) \in N \times N : 2x + y = 10\}$ and $R_2 = \{(x,y) \in N \times N : x + 2y = 10\}$. Then

In the set $N$,a relation $R$ is defined as $aRb \Leftrightarrow b$ is divisible by $a$. Then $R$ is:

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