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December 9th, 2016, 10:35 AM   #1
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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.
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December 9th, 2016, 02:48 PM   #2
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Originally Posted by Mathsisthebestandisamazin View Post
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.
For R to be reflexive we must have that {x,x} is in R. Since we have {a,a}, {b,b}, {c,c}, {d,d}, {e,e} in R we can say R is reflexive. We can't say that {a,a} is reflexive because this is just a single element of R.

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
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Last edited by topsquark; December 9th, 2016 at 02:50 PM.
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