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

Reflexive and symmetric but not transitive

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.