Question

(a) Let A = {0,1,2}. R = {(0,0), (0,1), (0,2), (1,1), (1,2), (2,2)} and
S = {(0,0), (1,1), (2,2)} be two relations on A.
(i) Show that R is a partial order relation.
(ii) Is R a total order relation?
(iii) Show that S is an equivalence relation.
1

Answer 1

1) Since $(0,0) ,(1,1),(2,2) \in R$

Therefore $R$ is reflexive .

$R$ is antisymmetric if $(x,y)\in R \ and \ (y,x)\in R \implies x=y$

Therefore $R$ is antisymmetric .

$R$ is transitive if for $x,y,z \in A$ , $(x,y)\in R \ and \ (y,z)\in R \implies (x,z)\in R$

Therefore $R$ is transitive.

Hence $R$ is a partial order relation.

2) $R$ is called total order relation if for any $x,y \in A , either \ (x,y)\in R \ or \ (y,x)\in R$

As any two elements of $A$ are Related , therefore $R$ is total order.

3) As (0,0),(1,1),(2,2) $\in S$ , Therefore $S$

is reflexive.

$S$ is symmetric if $(x,y)\in S \implies (y,x)\in S$

Therefore $S$ is symmetric.

$S$ is transitive if $(x,y)\in S \ and \ (y,z) \in S \implies (x,z)$

Clearly $S$ is transitive.

Therefore $S$ is a eqivalence relation.