Question #245937
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)
1
Expert's answer
2022-02-14T17:36:07-0500

Let us determine if the following is an equivalence relation on X={1,2,3,4,5}.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)}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)}R,\{(1, 1), (2, 2), (3, 3), (4. 4), (5. 5)\}\subset R, we conclude that the relation is reflexive. Since R1={(1,1),(2,2),(3,3),(4.4),(5.5),(3,1),(1,3)}=R,R^{-1}=\{(1, 1), (2, 2), (3, 3), (4. 4), (5. 5), (3, 1), (1, 3)\}=R, the relation RR is symmetric. Taking into account that (a,b),(b,c)R(a,b),(b,c)\in R implies (a,c)R(a,c)\in R for all (a,b),(b,c)R,(a,b),(b,c)\in R, we conclude that this relation is transitive.

Let us enumerate its equivalence classes. Recall that [a]={xX:(a,x)R}.[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}.[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)}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)}R,\{(1, 1), (2, 2), (3, 3), (4. 4), (5. 5)\}\subset R, we conclude that the relation is reflexive. Since R1={(1,1),(2,2),(3,3),(4,4),(5,5),(5,1),(1,5),(5,3),(3,5),(3,1),(1,3)}=R,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 RR is symmetric. Taking into account that (a,b),(b,c)R(a,b),(b,c)\in R implies (a,c)R(a,c)\in R for all (a,b),(b,c)R,(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}.[1]=\{1,3,5\},\ [2]=\{2\},\ [4]=\{4\}.


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


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


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