Is R1 an Equivalence Relation or Partial Order on Set M?

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In summary: It is not symmetric because the matrix is not symmetric about the main diagonal. It is also not transitive because there are 0s in the same position in both matrices, indicating that the composition of relations is not always in R1. Therefore, R1 is not an equivalence relation.
  • #1
Yusuke_Ichigo
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R1 = {(p,q) | p-q is even}; R1 on set M = {1,2,3,4};
Ordering of M: 1,2,3,4

a) Write the relation R1 as a set of ordered pairs. Draw the arrow diagram and determine whether R1 are functions or not. Explain your answer.

b) Hence, determine whether the relation R1 is an equivalence relation or partial order (or neither both).

c) Describe how can the digraph of the relation R1 be used to determine whether R1 is an equivalence relation. Your answer should include the digraph and detail description.

d) Determine the matrix of the relation R1 (relative to the given orderings). Now, reorder R1 as 3,2,1,4, thus determine the new matrix obtained.

e) Another technique to test for reflexive, symmetric and transitivity is by using the matrix of relation. Analyze matrix of the relation R1 to determine whether R1 is an equivalence relation.

I only understand this R1 on set M = {(1,1) , (1,3) , (2,2) , (2,4) , (3,1) , (3,3) , (4,2) , (4,4)}. Can anyone help?
Thank you.
 
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  • #2


a) R1 = {(1,1), (1,3), (2,2), (2,4), (3,1), (3,3), (4,2), (4,4)}
The arrow diagram for R1 would look like this:

1 → 1, 3
2 → 2, 4
3 → 1, 3
4 → 2, 4

This relation is not a function because the element 1 in the domain is related to two different elements, 1 and 3, in the range. A function can only have one output for each input.

b) The relation R1 is neither an equivalence relation nor a partial order. It is not reflexive because not all elements in M are related to themselves. It is not symmetric because if (p,q) is in R1, then (q,p) is not necessarily in R1. It is not transitive because if (p,q) is in R1 and (q,r) is in R1, then (p,r) is not necessarily in R1.

c) The digraph of R1 can be used to determine whether R1 is an equivalence relation by checking if it satisfies the properties of reflexivity, symmetry, and transitivity. A digraph for R1 would look like this:

1 → 1, 3
2 → 2, 4
3 → 1, 3
4 → 2, 4

From this digraph, we can see that R1 is not reflexive because elements 1 and 4 are not connected to themselves. It is also not symmetric because there are arrows going in only one direction. Lastly, it is not transitive because there is no connection between elements 1 and 4, even though they are both related to element 3.

d) The matrix for R1 would be:
1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1

Reordering R1 as 3, 2, 1, 4 would result in the new matrix:
1 0 1 0
1 0 1 0
0 1 0 1
0 1 0 1

e) From the matrix, we can see that R1 is not reflexive because there are 0s on
 
  • #3


a) The relation R1 can be written as a set of ordered pairs as follows: {(1,1), (1,3), (2,2), (2,4), (3,1), (3,3), (4,2), (4,4)}. The arrow diagram would have 4 arrows pointing from 1 to 1, 3 to 1, 2 to 2, and 4 to 2. This relation is not a function because there are two different inputs (1 and 3) that have the same output (1). In a function, each input must have a unique output.

b) The relation R1 is neither an equivalence relation nor a partial order. It is not reflexive because (1,1) and (2,2) are the only pairs that satisfy the condition p-q is even. It is not symmetric because (1,3) is in R1 but (3,1) is not. It is not transitive because (1,3) and (3,1) are in R1, but (1,1) is not.

c) The digraph of the relation R1 can be used to determine whether R1 is an equivalence relation by checking if it is reflexive, symmetric, and transitive. A reflexive relation would have a loop at each vertex, a symmetric relation would have arrows going in both directions between vertices, and a transitive relation would have a path from one vertex to another through another vertex. In the case of R1, there is only one loop at (2,2), no arrows going in both directions, and no path from (1,3) to (1,1) through (3,1). Therefore, R1 is not an equivalence relation.

d) The matrix of the relation R1 would be:

1 0 1 0
0 1 0 1
1 0 1 0
0 1 0 1

Reordering R1 as 3,2,1,4 would result in the following matrix:

1 0 0 1
0 1 1 0
0 1 1 0
1 0 0 1

e) The matrix of the relation R1 can be used to determine whether R1 is an equivalence relation by checking if it is reflexive, symmetric, and transitive. A reflex
 

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