Answer to Question #245937 in Discrete Mathematics for Mico

Question

2. {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (1, 5).(5, 1), (3, 5). (5, 3), (1, 3), (3, 1))


Determine if the following is an equivalence relation on X = (1, 2, 3, 4, 5). If the following are equivalence relation, then enumerate its equivalence classes. 1. {(1, 1), (2, 2), (3, 3), (4. 4). (5. 5), (1, 3), (3, 1))

3. [(x, y) 13 divides x + y)

Answer 1

Let us determine if the following is an equivalence relation on $X = \{1, 2, 3, 4, 5\}.$ If the following are equivalence relation, then let us enumerate its equivalence classes.

1. $R=\{(1, 1), (2, 2), (3, 3), (4. 4), (5. 5), (1, 3), (3, 1)\}$

Taking into account that $\{(1, 1), (2, 2), (3, 3), (4. 4), (5. 5)\}\subset R,$ we conclude that the relation is reflexive. Since $R^{-1}=\{(1, 1), (2, 2), (3, 3), (4. 4), (5. 5), (3, 1), (1, 3)\}=R,$ the relation $R$ is symmetric. Taking into account that $(a,b),(b,c)\in R$ implies $(a,c)\in R$ for all $(a,b),(b,c)\in R,$ we conclude that this relation is transitive.

Let us enumerate its equivalence classes. Recall that $[a]=\{x\in X:(a,x)\in R\}.$ It follows that there are 4 different equivalence classes: $[1]=\{1,3\},\ [2]=\{2\},\ [4]=\{4\},\ [5]=\{5\}.$

2. $R=\{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (1, 5),(5, 1), (3, 5), (5, 3), (1, 3), (3, 1)\}$

Taking into account that $\{(1, 1), (2, 2), (3, 3), (4. 4), (5. 5)\}\subset R,$ we conclude that the relation is reflexive. Since $R^{-1}=\{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (5, 1), (1, 5),(5, 3), (3, 5), (3, 1),(1, 3)\}=R,$

the relation $R$ is symmetric. Taking into account that $(a,b),(b,c)\in R$ implies $(a,c)\in R$ for all $(a,b),(b,c)\in R,$ we conclude that this relation is transitive.

Let us enumerate its equivalence classes. It follows that there are 3 different equivalence classes: $[1]=\{1,3,5\},\ [2]=\{2\},\ [4]=\{4\}.$

3. $R=\{(x, y) :13\text{ divides }x + y\}$

Taking into account that 13 does not divide $2=1+1,$ we conclude that $(1,1)\notin R,$ and hence $R$ is not reflexive. Therefore, $R$ is not an equivalence relation.