Question

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

• A. Only I is true
• B. Only II is true
• C. Both I and II are true
• D. Neither I nor II is true

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
A relation $R$ on set $A$ is a subset of $A \times A$. For $R$ to be an equivalence relation, it must be reflexive ($aRa$), symmetric ($aRb \implies bRa$), and transitive ($aRb$ and $bRc \implies aRc$).
Step 2: Key Formula or Approach:
Identify all pairs $(a, b)$ from $S \times S$ where $S = \{-2, -1, 0, 1, 2\}$ such that $ab>-1$.
Step 3: Detailed Explanation:
1. Counting elements (I): Total possible pairs in $S \times S$ is $5 \times 5 = 25$.
Pairs failing $1+ab>0$ (i.e., $ab \leq -1$): $(-2, 1), (-2, 2), (-1, 1), (-1, 2), (1, -2), (1, -1), (2, -2), (2, -1)$.
Total failing pairs = 8. Number of elements in $R = 25 - 8 = 17$. So, Statement I is true. 2. Equivalence Check (II): - Reflexive: $1 + a^2>0$ is true for all $a \in S$.
- Symmetric: $1 + ab>0 \implies 1 + ba>0$. True.
- Transitive: Let $a=2, b=0, c=-2$.
$1 + (2)(0) = 1>0 \implies (2, 0) \in R$.
$1 + (0)(-2) = 1>0 \implies (0, -2) \in R$.
But $1 + (2)(-2) = -3 \ngtr 0 \implies (2, -2) \notin R$.
Since transitivity fails, $R$ is not an equivalence relation. So, Statement II is false.
Step 4: Final Answer:
Only statement I is true.