Practicing Success
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 transitive reflexive and symmetric symmetric and transitive an equivalence relation |
reflexive and symmetric |
For any x e Z, we have $∴|x-x|≤1$ 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. |