Relation R Analysis

Set Definition and Condition

The set is defined as $S = \{-2, -1, 0, 1, 2\}$.

The relation R on $S \times S$ is defined by the condition $(a, b) \in R$ if and only if $1 + ab > 0$. We need to check two statements about this relation.

Statement I Verification: Number of Elements

We systematically list all pairs $(a, b)$ from $S \times S$ where $1 + ab > 0$. The total number of pairs in $S \times S$ is $5 \times 5 = 25$. Let's check the condition for each element $a \in S$:

• If $a = -2$: Pairs are $(-2, -2), (-2, -1), (-2, 0)$. (3 pairs)
• If $a = -1$: Pairs are $(-1, -2), (-1, -1), (-1, 0)$. (3 pairs)
• If $a = 0$: Pairs are $(0, -2), (0, -1), (0, 0), (0, 1), (0, 2)$. (5 pairs)
• If $a = 1$: Pairs are $(1, 0), (1, 1), (1, 2)$. (3 pairs)
• If $a = 2$: Pairs are $(2, 0), (2, 1), (2, 2)$. (3 pairs)

The total number of elements in R is the sum of pairs found: $3 + 3 + 5 + 3 + 3 = 17$. Therefore, Statement I is true.

Statement II Verification: Equivalence Relation Properties

For R to be an equivalence relation, it must be reflexive, symmetric, and transitive.

Reflexivity

Check if $(a, a) \in R$ for all $a \in S$. This requires $1 + a \cdot a > 0$, which is $1 + a^2 > 0$. Since $a^2 \ge 0$ for any real number $a$, $1 + a^2$ is always greater than 0. Thus, R is reflexive.

Symmetry

Check if $(a, b) \in R$ implies $(b, a) \in R$. If $1 + ab > 0$, does $1 + ba > 0$? Since multiplication is commutative ($ab = ba$), the condition $1 + ab > 0$ is equivalent to $1 + ba > 0$. Thus, R is symmetric.

Transitivity

Check if $(a, b) \in R$ and $(b, c) \in R$ implies $(a, c) \in R$. Let's test this:

• Consider the pair $(a, b) = (-2, 0)$. We have $1 + (-2)(0) = 1 > 0$, so $(-2, 0) \in R$.
• Consider the pair $(b, c) = (0, 1)$. We have $1 + (0)(1) = 1 > 0$, so $(0, 1) \in R$.
• Now check the pair $(a, c) = (-2, 1)$. The condition is $1 + (-2)(1) = 1 - 2 = -1$. Since $-1 \ngtr 0$, the pair $(-2, 1)$ is not in R.

Since we found a case where $(a, b) \in R$ and $(b, c) \in R$ but $(a, c) \notin R$, the relation R is not transitive.

Because R is not transitive, it is not an equivalence relation. Statement II is false.

Conclusion on Statements

Based on the analysis, only Statement I is true, while Statement II is false.