Solution to 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. 1.Show that R is a … - Sikademy
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Archangel Macsika

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. 1.Show that R is a partial order relation 2.Is R a total order relation? 3.Show that S is an equivalence relation.

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1.1 Since \{(0,0),(1,1),(2,2)\}\subset R , then R is reflexive.

1.2 For each pair (a,b)\in R if (b,a)\in R then a=b , hence R is a symmetric relation.

1.3 If (a,b)\in R and (b,c)\in R , then (a,c)\in R , hence R is transitive.

By 1.1, 1.2 and 1.3 R is a partial order relation.

It is easy to see that R is natural order (\leq) relation on the set A=\{0,1,2\}, hence (1) R is a partial order relation and R is total.

2. R is total.

(a,b)\in S if and only if a=b , hence

3.1 Since \{(0,0),(1,1),(2,2)\}\subset S, then S is reflexive.

3.2 If (a,b)\in S , then a=b and (b,a)\in S , hence S is a symmetric relation.

3.3 If (a,b)\in S and (b,c)\in S, then a=b=c and (a,c)\in S , hence S is transitive.

By 3.1, 3.2 and 3.3 S is an equivalence relation.

1.1 Since \{(0,0),(1,1),(2,2)\}\subset R , then R is reflexive.

1.2 For each pair (a,b)\in R if (b,a)\in R then a=b , hence R is a symmetric relation.

1.3 If (a,b)\in R and (b,c)\in R , then (a,c)\in R , hence R is transitive.

By 1.1, 1.2 and 1.3 R is a partial order relation.

It is easy to see that R is natural order (\leq) relation on the set A=\{0,1,2\}, hence (1) R is a partial order relation and R is total.

2. R is total.

(a,b)\in S if and only if a=b , hence

3.1 Since \{(0,0),(1,1),(2,2)\}\subset S, then S is reflexive.

3.2 If (a,b)\in S , then a=b and (b,a)\in S , hence S is a symmetric relation.

3.3 If (a,b)\in S and (b,c)\in S, then a=b=c and (a,c)\in S , hence S is transitive.

By 3.1, 3.2 and 3.3 S is an equivalence relation.

1.1 Since \{(0,0),(1,1),(2,2)\}\subset R , then R is reflexive.

1.2 For each pair (a,b)\in R if (b,a)\in R then a=b , hence R is a symmetric relation.

1.3 If (a,b)\in R and (b,c)\in R , then (a,c)\in R , hence R is transitive.

By 1.1, 1.2 and 1.3 R is a partial order relation.

It is easy to see that R is natural order (\leq) relation on the set A=\{0,1,2\}, hence (1) R is a partial order relation and R is total.

2. R is total.

(a,b)\in S if and only if a=b , hence

3.1 Since \{(0,0),(1,1),(2,2)\}\subset S, then S is reflexive.

3.2 If (a,b)\in S , then a=b and (b,a)\in S , hence S is a symmetric relation.

3.3 If (a,b)\in S and (b,c)\in S, then a=b=c and (a,c)\in S , hence S is transitive.

By 3.1, 3.2 and 3.3 S is an equivalence relation.

1.1 Since \{(0,0),(1,1),(2,2)\}\subset R , then R is reflexive.

1.2 For each pair (a,b)\in R if (b,a)\in R then a=b , hence R is a symmetric relation.

1.3 If (a,b)\in R and (b,c)\in R , then (a,c)\in R , hence R is transitive.

By 1.1, 1.2 and 1.3 R is a partial order relation.

It is easy to see that R is natural order (\leq) relation on the set A=\{0,1,2\}, hence (1) R is a partial order relation and R is total.

2. R is total.

(a,b)\in S if and only if a=b , hence

3.1 Since \{(0,0),(1,1),(2,2)\}\subset S, then S is reflexive.

3.2 If (a,b)\in S , then a=b and (b,a)\in S , hence S is a symmetric relation.

3.3 If (a,b)\in S and (b,c)\in S, then a=b=c and (a,c)\in S , hence S is transitive.

By 3.1, 3.2 and 3.3 S is an equivalence relation.

1.1 Since \{(0,0),(1,1),(2,2)\}\subset R , then R is reflexive.

1.2 For each pair (a,b)\in R if (b,a)\in R then a=b , hence R is a symmetric relation.

1.3 If (a,b)\in R and (b,c)\in R , then (a,c)\in R , hence R is transitive.

By 1.1, 1.2 and 1.3 R is a partial order relation.

It is easy to see that R is natural order (\leq) relation on the set A=\{0,1,2\}, hence (1) R is a partial order relation and R is total.

2. R is total.

(a,b)\in S if and only if a=b , hence

3.1 Since \{(0,0),(1,1),(2,2)\}\subset S, then S is reflexive.

3.2 If (a,b)\in S , then a=b and (b,a)\in S , hence S is a symmetric relation.

3.3 If (a,b)\in S and (b,c)\in S, then a=b=c and (a,c)\in S , hence S is transitive.

By 3.1, 3.2 and 3.3 S is an equivalence relation.


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