NOTE: This is a multi-part . Once an answer is submitted, you will be unable to return to this part Determine whether the relation on the set of all people is reflexive, symmetric, antisymmetric, and/or transitive, where (a. b) E Rif and only if ind b have a common grandparent. (Check all that apply) reflexive symmetric transitive antisymmetric
Accepted Solution
A:
Answer:A. This relation is Reflexive and SymmetricStep-by-step explanation:A. Before starting to test this relation, letΒ΄s remind the definition of these properties. Let R be a relation defined as RβAΓA, where A is an arbitrary set:R is Reflexive β (βa β A)((a,a) β R)R is Symmetric β (βa,b β A)("(a,b) β R"β"(b,a) β R")R is Anti-symmetric β (βa,b β A)("(a,b) β R"β§"(b,a) β R" β a=b) R is Transitive β Β (βa,b,c β A)("(a,b) β R"β§"(b,c) β R" β "(a,c) β R") Where (a,b)βR β a is Related to bThe question defines R as follows: "a is Related to b" β a and b have a common grandparent. We use the definition and some logic to prove whether R fulfill the definitions above or not:R is Reflexive: (we try to show that Β (βa β A)("(a,a) β R") where A is the set of people we work on)the elements are people, so we suppose that a has a grandparent. Because a has the same grandparent as a (they are the same person, so they have the same family) and it happens for all people you consider. We conclude that a is Related to a ((a,a) β R), and because a was arbitrary we also shown that this happens to everyone (βa β A)("(a,a) β R"). This proves that R is Reflexive.R is Symmetric: (Remember the definition above)Suppose (a,b) β R, then, "a has a common grandparent with b" and because the grandparent that they share is fixed, itΒ΄s true that "b has a common grandparent with a" so we have (b,a) β R. Because we are able to do this whenever we have (a,b) β R we conclude Β (β(a,b) β A)("(a,b) β R"β"(b,a) β R") Β and that R is Symmetric.R is Anti-symmetric: (R does not satisfy this definition)ItΒ΄s valid to give an example that does not fulfill the definition to prove definitely that R is not Anti-symmetric. Suppose a and b are brothers, because of this (a,b)βR, and (b,a)βR (because we proved that R is symmetric) but aβ b. Now we now that ("(a,b) β R"β§"(b,a) β R" β a=b) does not happen to every pair in R, so we conclude that R is not Anti-symmetric.R is transitive: (R does not satisfy this definition either)Suppose (a,b) β R and (b,c) β R, a and b have a grandparent in common, and the same happens with b and c. ItΒ΄s not hard to think that, for example, a has the same grandfather as b but b has the same grandmother with c. We show that a could not necessarily be related to c. so (βa,b,c β A)("(a,b) β R"β§"(b,c) β R" β "(a,c) β R") itΒ΄s not true, and R is not Transitive.