Consider the relation R on the set {-2, -1, 0 | vedclass

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(A) Let the set be $S = \{-2, -1, 0, 1, 2\}$. The total number of possible ordered pairs $(a, b)$ is $5 	imes 5 = 25$.We need to find pairs such that

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Only $I$ is true

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(A) Let the set be $S = \{-2, -1, 0, 1, 2\}$. The total number of possible ordered pairs $(a, b)$ is $5 	imes 5 = 25$.We need to find pairs such that $1 + ab > 0$,which is equivalent to $ab > -1$.It is easier to count the pairs where $ab \leq -1$ and subtract from $25$.The pairs $(a, b)$ where $ab \leq -1$ are:$(-2, 1) \Rightarrow ab = -2$$(1, -2) \Rightarrow ab = -2$$(-2, 2) \Rightarrow ab = -4$$(2, -2) \Rightarrow ab = -4$$(-1, 1) \Rightarrow ab = -1$$(1, -1) \Rightarrow ab = -1$$(-1, 2) \Rightarrow ab = -2$$(2, -1) \Rightarrow ab = -2$There are $8$ such pairs. Thus,the number of elements in $R$ is $25 - 8 = 17$. Statement $I$ is true.For Statement $II$,check if $R$ is an equivalence relation. $A$ relation is an equivalence relation if it is reflexive,symmetric,and transitive.Reflexivity: $1 + a^2 > 0$ is true for all $a \in S$. So,it is reflexive.Symmetry: $1 + ab > 0 \iff 1 + ba > 0$. So,it is symmetric.Transitivity: Consider $(1, 0) \in R$ because $1 + 0 = 1 > 0$,and $(0, -2) \in R$ because $1 + 0 = 1 > 0$. However,$(1, -2) 
otin R$ because $1 + (1)(-2) = -1 
gtr 0$. Thus,$R$ is not transitive. Statement $II$ is false.

Consider the relation $R$ on the set $\{-2, -1, 0, 1, 2\}$ defined by $(a, b) \in R$ if and only if $1 + ab > 0$. Then,among the statements:
$I$. The number of elements in $R$ is $17$
$II$. $R$ is an equivalence relation

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Consider the relation $R$ on the set $\{-2, -1, 0, 1, 2\}$ defined by $(a, b) \in R$ if and only if $1 + ab > 0$. Then,among the statements:$I$. The number of elements in $R$ is $17$$II$. $R$ is an equivalence relation