Consider the relation R on the set -2,-1,0,1,2 defined by (a, b) ∈ R if and only if 1+ab > 0. Then, among the statements: I. The number of elements…

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

For Statement I, the relation is defined as $(a, b) \in R \iff 1 + ab > 0$(a,b)∈R⟺1+ab>0.

The pairs $(a, b)$(a,b) that do not belong to $R$R satisfy $1 + ab \le 0 \Rightarrow ab \le -1$1+ab≤0⇒ab≤−1.

The pairs satisfying $ab \le -1$ab≤−1 are $(-2, 1), (-2, 2), (-1, 1), (-1, 2), (1, -2), (1, -1), (2, -2), (2, -1)$(−2,1),(−2,2),(−1,1),(−1,2),(1,−2),(1,−1),(2,−2),(2,−1).

There are $8$8 such pairs.

The total number of pairs in $A \times A$A×A is $5 \times 5 = 25$5×5=25.

The number of elements in $R$R is $25 - 8 = 17$25−8=17.

Thus, Statement I is true.

For Statement II, checking transitivity gives $(-2, 0) \in R$(−2,0)∈R since $1 + (-2)(0) = 1 > 0$1+(−2)(0)=1>0.

Also, $(0, 2) \in R$(0,2)∈R since $1 + (0)(2) = 1 > 0$1+(0)(2)=1>0.

However, $(-2, 2) \notin R$(−2,2)∈/R since $1 + (-2)(2) = -3 \le 0$1+(−2)(2)=−3≤0.

Since $R$R is not transitive, it is not an equivalence relation.

Thus, Statement II is false.

Answer: Only I is true