R is a relation on the set Z of integers and it is given by (x, y) ∈R…

For any x e Z, we have

$|x-x|=0≤1$

$∴|x-x|≤1$ for all $x ∈ Z$

$⇒(x,x)∈R$ for all $x ∈ Z$

⇒ R is reflexive on Z.

Let $(x, y) ∈ R$. Then,

$|x-y|≤1⇒|y-x|≤1 ⇒(y, x) ∈R$

Thus, $(x, y) ∈ R⇒(y, x) ∈ R$.

So, R is a symmetric relation on Z.

We observe that $(1, 0) ∈ R$ and $(0, -1) ∈ R$, but $(1, -1) ∉ R$.

So, R is not a transitive relation on Z.