Question on reflexivity, symmetry, and transitivity (Relation on X (Attempt inside)?

  • Thread starter sam0617
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  • #1
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Let X = { a, b, c }

X x X = { (a,a), (b,b), (c,c) }
{ (a,b), (b,a), (a,c), (c,a) }
{ (b,c), (c,b) }

1. Symmetric but not reflexive or transitive:
R = { (a,b), (b,a), (a,a), (b,c), (c,b) }
How come this is right? Isn't aRb, bRa imply aRa? isn't that transitive? is it because (b,c,), (c,b) is there but not (b,b) the reason why R is not transitive?

I ask because the 2nd question is confusing. Here it is:
2. Symmetric and transitive but not reflexive:
R= { (a,a), (a,b), (b,a), (b,b) }
See how aRb, bRa implies aRa so therefore it's transitive? How come it doesn't hold for the 1st question??

Thank you for any help.
 

Answers and Replies

  • #2
micromass
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Transitive means that for ALL x,y,z:

[tex]xRy~\text{and}~yRz~\Rightarrow~xRz[/tex]

It must holds for ALL.

For the first one, it isn't transitive since if you take x=z=a and y=x, then you see that the above is not satisfied. So it doesn't hold for ALL x,y,z. It does hold for some x,y,z. But it does hold for some. But some isn't enough to imply transitivity.

In (2), it does hold for ALL, so it is transitive.
 
  • #3
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Transitive means that for ALL x,y,z:

[tex]xRy~\text{and}~yRz~\Rightarrow~xRz[/tex]

It must holds for ALL.

For the first one, it isn't transitive since if you take x=z=a and y=x, then you see that the above is not satisfied. So it doesn't hold for ALL x,y,z. It does hold for some x,y,z. But it does hold for some. But some isn't enough to imply transitivity.

In (2), it does hold for ALL, so it is transitive.

I'm sorry, I don't understand what the x=z=a and y=x then it wouldn't satisfy above.
Could you explain more?

EDIT: Then to make question 1 transitive, all I would have to add is (b,b) ?
 
  • #4
micromass
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Sorry, typo. I meant that if x=z=b and y=c, then it isn't true that bRc and cRb and bRb.

Adding (b,b) would not make it transtive.

Indeed, we also don't have

aRb and bRc => aRc

since (a,c) is not in the relation.
 

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