Solution to (a) Let A = {0,1,2}. R = {(0,0), (0,1), (0,2), (1,1), (1,2), (2,2)} and S … - Sikademy
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Archangel Macsika

(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

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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.

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