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:
We are given a relation \( R \) defined on a finite set \( S = \{-2, -1, 0, 1, 2\} \) such that \((a, b) \in R \iff 1 + ab>0\).
We need to determine the total number of ordered pairs in \( R \) to verify Statement I, and check if \( R \) is reflexive, symmetric, and transitive to verify Statement II.
Step 2: Key Formula or Approach:
An equivalence relation must satisfy three properties:
1. Reflexive: \((a, a) \in R\) for all \(a \in S\).
2. Symmetric: \((a, b) \in R \implies (b, a) \in R\).
3. Transitive: \((a, b) \in R\) and \((b, c) \in R \implies (a, c) \in R\).
To find the number of elements, we systematically check all pairs \((a,b)\).
Step 3: Detailed Explanation:
First, let's find the number of elements in \( R \).
The condition is \( 1 + ab>0 \implies ab>-1 \implies ab \ge 0 \) (since \( a, b \) are integers).
Let's check for each \( a \in \{-2, -1, 0, 1, 2\} \):
If \( a = -2 \): \( b \) can be \( -2, -1, 0 \). (3 pairs: \((-2,-2), (-2,-1), (-2,0)\)).
If \( a = -1 \): \( b \) can be \( -2, -1, 0 \). (3 pairs: \((-1,-2), (-1,-1), (-1,0)\)).
If \( a = 0 \): \( ab = 0 \ge 0 \) is true for all \( b \in \{-2, -1, 0, 1, 2\} \). (5 pairs).
If \( a = 1 \): \( b \) can be \( 0, 1, 2 \). (3 pairs: \((1,0), (1,1), (1,2)\)).
If \( a = 2 \): \( b \) can be \( 0, 1, 2 \). (3 pairs: \((2,0), (2,1), (2,2)\)).
Total number of elements in \( R = 3 + 3 + 5 + 3 + 3 = 17 \).
Thus, Statement I is true.
Now let's check if \( R \) is an equivalence relation.
- Reflexive: \( 1 + a^2>0 \) is true for all \(a\), so it is reflexive.
- Symmetric: \( 1 + ab>0 \implies 1 + ba>0 \), so it is symmetric.
- Transitive: Let \( a = -1, b = 0, c = 1 \).
We have \((-1, 0) \in R\) (since \(1+0>0\)) and \((0, 1) \in R\) (since \(1+0>0\)).
However, \((-1, 1) \notin R\) because \( 1 + (-1)(1) = 0 \ngtr 0 \).
Since it is not transitive, it is not an equivalence relation.
Thus, Statement II is false.
Step 4: Final Answer:
Only Statement I is true.