The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Equivalnce Relation

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.
Reply 1
Avatar for Zacken
Zacken
22
Original post by Deeboss
FYI: Sorry for the lengthy post just want to put my thoughts to paper, so people dont think I am here for quick answers.


No worries, good to see some work for once. The relation is reflexive, like you said, because every element is related to itself. It is symmetric as well, like you said again. But it is also transitive, the example you gave involving b and d not being transitive doesn't work.

The only relations involving b and d are (b,b), (d,d), (b,d) and (d,b). But this is clearly transitive, since b~b and b~d implies b~d which is true. b~d and d~b implies b~b, again true, etc...
Reply 2
Avatar for Deeboss
Deeboss
OP
11
Original post by Zacken
No worries, good to see some work for once. The relation is reflexive, like you said, because every element is related to itself. It is symmetric as well, like you said again. But it is also transitive, the example you gave involving b and d not being transitive doesn't work.

The only relations involving b and d are (b,b), (d,d), (b,d) and (d,b). But this is clearly transitive, since b~b and b~d implies b~d which is true. b~d and d~b implies b~b, again true, etc...


Thank you very much for the help sir. I understand where I went wrong.
Reply 3
Avatar for Zacken
Zacken
22
Original post by Deeboss
Thank you very much for the help sir. I understand where I went wrong.


No worries

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you