Assume that R is a relation on the set Z of integers and it is given by $(x, y) ∈R ⇔|x − y| ≤ 1$. Then, R is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → Reflexive and symmetric but not transitive
Given relation:
$R = \{(x, y) \in \mathbb{Z} \times \mathbb{Z} : |x - y| \le 1\}$
Reflexive:
For any $x \in \mathbb{Z}$, $|x - x| = 0 \le 1$.
⇒ Reflexive ✔
Symmetric:
If $|x - y| \le 1$, then $|y - x| \le 1$ (since absolute value is symmetric).
⇒ Symmetric ✔
Transitive:
Suppose $|x - y| \le 1$ and $|y - z| \le 1$.
Then $|x - z| \le |x - y| + |y - z| \le 2$, which may be 2 (not ≤ 1).
Example: $x = 1$, $y = 2$, $z = 3$ → $|1 - 2| = 1$, $|2 - 3| = 1$, but $|1 - 3| = 2 \text{[ greater than ]} 1$.
⇒ Not transitive ✖
Therefore: Relation $R$ is reflexive and symmetric but not transitive.
Correct option: Reflexive and symmetric but not transitive.