how that R is an equivalence relation, and find all equivalence classes.(a) R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (1, 5), (5, 1), (3, 5), (5, 3), (1, 3), (3, 1)} on {1, 2, 3, 4, 5}(b) R = {(x, y) : 3 divides (x − y)} on R(c) R = {(x, y) : 3 divides (x + 2y)} on Z(d) R = {(x, y) : x and y are both even} on {1, 2, 3, 4, 5}
Question
how that R is an equivalence relation, and find all equivalence classes.(a) R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (1, 5), (5, 1), (3, 5), (5, 3), (1, 3), (3, 1)} on {1, 2, 3, 4, 5}(b) R = {(x, y) : 3 divides (x − y)} on R(c) R = {(x, y) : 3 divides (x + 2y)} on Z(d) R = {(x, y) : x and y are both even} on {1, 2, 3, 4, 5}
Solution
(a) An equivalence relation must satisfy three properties: reflexivity, symmetry, and transitivity.
Reflexivity: Every element is related to itself. This is true for R as (1,1), (2,2), (3,3), (4,4), (5,5) are in R.
Symmetry: If (a,b) is in R, then (b,a) must also be in R. This is also true for R as whenever (a,b) is in R, (b,a) is also in R.
Transitivity: If (a,b) and (b,c) are in R, then (a,c) must also be in R. This is true for R as whenever (a,b) and (b,c) are in R, (a,c) is also in R.
Therefore, R is an equivalence relation. The equivalence classes are: [1] = {1,3,5}, [2] = {2}, [3] = {1,3,5}, [4] = {4}, [5] = {1,3,5}.
(b) The relation R = {(x, y) : 3 divides (x − y)} on R is an equivalence relation because it is reflexive (3 divides x-x for all x), symmetric (if 3 divides x-y, then 3 divides y-x), and transitive (if 3 divides x-y and y-z, then 3 divides x-z). The equivalence classes are all real numbers that differ by a multiple of 3.
(c) The relation R = {(x, y) : 3 divides (x + 2y)} on Z is not an equivalence relation because it is not symmetric. For example, if x=1 and y=1, then 3 divides 1+21=3, but 3 does not divide 1+21=3. Therefore, this relation is not an equivalence relation and does not have equivalence classes.
(d) The relation R = {(x, y) : x and y are both even} on {1, 2, 3, 4, 5} is not an equivalence relation because it is not reflexive. For example, 1 is not related to itself because it is not even. Therefore, this relation is not an equivalence relation and does not have equivalence classes.
Similar Questions
Define equivalence relation on a set. Let R be a relation on the set of integers defined by aRb iff a − bis multiple of 5. Prove that R is equivalence relation.
Suppose a relation R = {(3, 3), (5, 5), (5, 3), (5, 5), (6, 6)} on S = {3, 5, 6}. Here R is known as _________a.equivalence relationb.symmetric relationc.transitive relation’’d.reflexive relat
Let R be the following equivalence relation on the set A = {1, 2, 3, 4, 5, 6}, R = {(1, 1),(2, 2),(3, 3),(4, 4),(5, 5),(6, 6),(1, 2),(2, 1),(2, 3),(3, 2),(1, 3),(3, 1),(5, 6),(6, 5)}. Find the partitions of A induced by R i.e. A/R.
etermine whether the each of the relation defined on the set of positive integers is reflexive,symmetric, antisymmetric, or transitive.(a) R = {(x, y) : xy = 2}(b) R = {(x, y) : xy ≥ 1}(c) R = {(x, y) : x = and2}(d) R = {(x, y) : 3 divides (x + 2and)}(It is) R = {(x, y) : x − and = 2}(f) R = {(x, y) : 3 divides (x − an
Given a set S = {1, 2, 3, 4, 5}, find the equivalence relation on S which generated by the partition{{1, 2}, {3}, {4, 5}}. Draw the graph of the relation.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.