Consider the following relations, where R is the set of real numbers and \(R_0\) is the set of non-zero real numbers.
Match List-I with List-II
| List-I | List-II | ||
| (A) | \(R_1 = \{(a, b) : ab \gt 0\}\) on \(R_0\) | (I) | Reflexive and symmetric but not transitive |
| (B) | \(R_2 = \{(a, b) : b - a \text{ is a non-negative integer}\}\) on R | (II) | An equivalence relation |
| (C) | \(R_3 = \{(a, b) : |a - b| \le 3\}\) on R | (III) | Symmetric but neither reflexive nor transitive |
| (D) | \(R_4 = \{(a, b) : a + b = 10\}\) on R | (IV) | Reflexive and transitive but not symmetric |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
(A) \(R_1 = \{(a, b) : ab \gt 0\}\) on \(R_0\)
This matches (II) An equivalence relation.
\(a \cdot a = a^{2} \gt 0\) for \(a \ne 0\), so reflexive
\(ab \gt 0\) gives \(ba \gt 0\), so symmetric
\(ab \gt 0\) means a and b have the same sign; if b and c also have the same sign, so do a and c, so transitive
(B) \(R_2 = \{(a, b) : b - a \text{ is a non-negative integer}\}\) on R
This matches (IV) Reflexive and transitive but not symmetric.
\(a - a = 0\), so reflexive
\(1 - 0 = 1\) but \(0 - 1 = -1\), so (0, 1) is in \(R_2\) and (1, 0) is not: not symmetric
\(c - a = (b - a) + (c - b)\) is a sum of two non-negative integers, so transitive
(C) \(R_3 = \{(a, b) : |a - b| \le 3\}\) on R
This matches (I) Reflexive and symmetric but not transitive.
\(|a - a| = 0 \le 3\), so reflexive
\(|b - a| = |a - b|\), so symmetric
(0, 3) and (3, 6) are in \(R_3\), but \(|0 - 6| = 6\), so (0, 6) is not: not transitive
(D) \(R_4 = \{(a, b) : a + b = 10\}\) on R
This matches (III) Symmetric but neither reflexive nor transitive.
\(1 + 1 \ne 10\), so not reflexive
\(a + b = 10\) gives \(b + a = 10\), so symmetric
(1, 9) and (9, 1) are in \(R_4\), but (1, 1) is not: not transitive