Consider the relation R on the set \ -2,-1,0,1,2\ defined by (a, b)...

The set is $A = \{-2, -1, 0, 1, 2\}$.

For Statement I, the relation is defined as $(a, b) \in R \iff 1 + ab > 0$.
The pairs $(a, b)$ that do not belong to $R$ satisfy $1 + ab \le 0 \Rightarrow ab \le -1$.
The pairs satisfying $ab \le -1$ are $(-2, 1), (-2, 2), (-1, 1), (-1, 2), (1, -2), (1, -1), (2, -2), (2, -1)$.
There are $8$ such pairs.
The total number of pairs in $A \times A$ is $5 \times 5 = 25$.
The number of elements in $R$ is $25 - 8 = 17$.
Thus, Statement I is true.

For Statement II, checking transitivity gives $(-2, 0) \in R$ since $1 + (-2)(0) = 1 > 0$.
Also, $(0, 2) \in R$ since $1 + (0)(2) = 1 > 0$.
However, $(-2, 2) \notin R$ since $1 + (-2)(2) = -3 \le 0$.
Since $R$ is not transitive, it is not an equivalence relation.
Thus, Statement II is false.

Answer: Only I is true