December 9th, 2016, 10:35 AM  #1 
Newbie Joined: Nov 2016 From: Texas Posts: 12 Thanks: 0  Equivalence Relations
Let S={a,b,c,d,e} and R= {(a,a),(a,c),(a,e),(b,b),(b,d),( c,a), (c,c),(c,e),(d,b),(d,d),(e,a),(e ,c),(e,e)} Is R is reflexive, symmetric, transitive and an equivalence relation ? TRUE/FALSE? My Method/Knowledge/Answer:  Knowledge Ok so I know the three classes and their rules are:  Reflexive = a~b Symmetric = a~b and b~a Transitive = If a~b and b~c then a~c. Method For example with R={(a,a)} = This would be Reflexive (Because a is equal to a) R={(a,c)}/{(c,a)} = This would be Symmetric (Because a is equal to b and b is equal to a) R={(a,e)}/{(e,a)} = This would be Transitive (Because if a is equal to b and b is equal to c then a is equal to c) R={(c,c)} = This would be Reflexive (Because a is equal to a) R={(c,a)}/{(a,c)} = This would be Symmetric (Because a is equal to b and b is equal to a) R={(c,e)}/{(e,c)} = This would be Transitive (Because if a is equal to b and b is equal to c then a is equal to c) R={(e,e)} = This would be Reflexive (Because a is equal to a) R={(e,a)} = This would be Symmetric (Because a is equal to b and b is equal to a) R={(e,c)}/{(c,e)} = This would be Transitive (Because if a is equal to b and b is equal to c then a is equal to c) Now this is why I think the whole answer is false is this next part R={(b,b)} = This would be Reflexive (Because a is equal to a) R={(b,d)}/{(d,b)} = This would be Symmetric (Because a is equal to b and b is equal to a) For transitive their would be nothing? Am I right? (Just because their is no transitive here would it make the whole answer false?) FYI: Sorry for the lengthy post just want to put my thoughts to paper, so people dont think I am here for quick answers. 
December 9th, 2016, 02:48 PM  #2  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,205 Thanks: 901 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
The same goes for symmetry: If we have {x, y} we must have {y, x} in R. We don't have to have all such pairs but again we are looking for a subset of R such that this is true. We can't say that a~c is reflexive...this is only one element in the set defining R. The same kind of thing goes for transitive. We can't say that {c, e}, {e, c} in R defines a transitive element without also having an element b such that {c, b} and {b, d} and {c, d} are all in R. Your R is reflexive and symmetric. Take another look at the transitive property: If we have {a, c} and {c, e} then we must also have {a, e}. We have {b, d} and {d, e}. Do we have {b, e}? etc. To answer your direct question, just because we have {b, d} in R does not mean that there is a transitive element involved, as long as we don't have any elements (d, x) and no element (b, x). But it can have the element {b, d} alone. Dan Last edited by topsquark; December 9th, 2016 at 02:50 PM.  

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