Solution to {F} 2. Find the transitive closures of these relations on {1, 2, 3, 4}. a) … - Sikademy
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Archangel Macsika

{F} 2. Find the transitive closures of these relations on {1, 2, 3, 4}. a) {(1, 2), (2,1), (2,3), (3,4), (4,1)} b) {(2, 1), (2,3), (3,1), (3,4), (4,1), (4, 3)} c) {(1, 2), (1,3), (1,4), (2,3), (2,4), (3, 4)} d) {(1, 1), (1,4), (2,1), (2,3), (3,1), (3, 2), (3,4), (4, 2)} 3. Find the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is a) reflexive and transitive. b) symmetric and transitive. c) reflexive, symmetric, and transitive. 4. Which of these relations on {0, 1, 2, 3} are equivalence relations? Determine the properties of an equivalence relation that the others lack. a) {(0, 0), (1, 1), (2, 2), (3, 3)} b) {(0, 0), (0, 2), (2, 0), (2, 2), (2, 3), (3, 2), (3, 3)} c) {(0, 0), (1, 1), (1, 2), (2, 1), (2, 2), (3, 3)} d) {(0, 0), (1, 1), (1, 3), (2, 2), (2, 3), (3, 1), (3, 2),(3, 3)} e) {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0),(2, 2), (3, 3)}

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2.

a) {(1, 2), (2,1), (2,3), (3,4), (4,1)} contains an oriented cycle {(1, 2), (2,3), (3,4), (4,1)}, the transitive closure of which is the maximal relation:

{(1,1), (1, 2), (1,3), (1,4), (2,1), (2, 2), (2,3), (2,4), (3,1), (3, 2), (3,3), (3,4), (4,1), (4, 2), (4,3), (4,4)}

Therefore, the transitive closure of the initial relation will be also the maximal relation.


b) {(2, 1), (2,3), (3,1), (3,4), (4,1), (4, 3)}

The transitive closure of the relation {(2,3), (3,4), (4,1)} is the strong linear order on the set {1, 2, 3, 4}. The pairs (2, 1), (3,1) lie in the transitive closure and do not affect on the transitive closure.

But the pairs {(3,4), (4,3)} provide the equivalence of the elements 3 and 4. Therefore, the transitive closure of the initial relation is the following

{(2,3), (2,4), (2,1), (3, 4), (3,1), (4,3), (4,1)}


c) {(1, 2), (1,3), (1,4), (2,3), (2,4), (3, 4)}

The transitive closure of the relation {(1,2), (2,3), (3,4)} is the strong linear order on the set {1, 2, 3, 4}. All the other pairs {(1,3), (1,4), (2,4)} are consistent with the transitive closure. Therefore, the transitive closure of the initial relation is the following

{(1, 2), (1,3), (1, 4), (2,3), (2,4), (3,4)}


d) The relation {(1, 1), (1,4), (2,1), (2,3), (3,1), (3, 2), (3,4), (4, 2)} contains the cyclic relation

{(1,4), (4, 2), (2,3), (3,1)}, the transitive closure of which is the maximal relation:

{(1,1), (1, 2), (1,3), (1,4), (2,1), (2, 2), (2,3), (2,4), (3,1), (3, 2), (3,3), (3,4), (4,1), (4, 2), (4,3), (4,4)}

Therefore, the transitive closure of the initial relation will be also the maximal relation.


3. {(1, 2), (1, 4), (3, 3), (4, 1)}

a) the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is reflexive and transitive must contain the transitive closure of this relation and the diagonal relation, that is, the union \{(1, 2), (1, 4), (3, 3), (4, 1), (4,2)\}\cup \{(1,1), (2,2), (3,3), (4,4)\} = \{(1,1), (1, 2), (1, 4), (2,2), (3, 3), (4, 1), (4, 2), (4,4)\}

Since the latter relation is already reflexive and transitive, it is the required minimal such relation.


b) If a relation is symmetric and transitive, then it is also reflexive:

Indeed, if (x,y)\in R then (y,x)\in R (symmetry) and (x,x)\in R (transitivity).

Therefore, the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is symmetric and transitive, is the same as in c).


c) If a relation is reflexive, symmetric, and transitive, it is an equivalence relation. We have 4\sim 1\sim 2 and 3\sim 3. Therefore, the smallest relation containing the relation {(1, 2), (1, 4), (3, 3), (4, 1)} that is reflexive, symmetric, and transitive, is the following

{(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (4, 1), (4, 2), (4, 4), (3, 3)}.


4.

a) R= {(0, 0), (1, 1), (2, 2), (3, 3)}

b) R={(0, 0), (0, 2), (2, 0), (2, 2), (2, 3), (3, 2), (3, 3)}

c) R={(0, 0), (1, 1), (1, 2), (2, 1), (2, 2), (3, 3)}

d) R={(0, 0), (1, 1), (1, 3), (2, 2), (2, 3), (3, 1), (3, 2),(3, 3)}

e) R={(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 2), (3, 3)}

Fill the following table:

\begin{vmatrix}Relation & a & b & c & d & e\\ Reflexive & + & - & + & + & +\\ Symmetric & + & + & + & + & -\\ Transitive & + & - & + & - & -\\ \end{vmatrix}

Relation R in b) is not reflexive, since (1,1)\notin R

Relation R in e) is not symmetric, since (1,2)\in R but (2,1)\notin R.

Relation R in b) is not transitive, since (0,2), (2,3)\in R but (0, 3)\notin R.

Relation R in d) is not transitive, since (2,3),(3,1)\in R but (2, 1)\notin R.

Relation R in e) is not transitive, since (2, 0), (0, 1)\in R but (2, 1)\notin R.

Therefore, only the relations in a) and c) are equivalence relations.

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Question ID: mtid-5-stid-8-sqid-200-qpid-88