Is a Relation Transitive If No Complete Pair Set Exists for Verification?

Click For Summary
A relation R on a set S is considered transitive if the condition holds that (x, y) and (y, z) implies (x, z) for all pairs in R. If no complete pair set exists to verify this condition, the relation can still be deemed transitive, as the absence of counterexamples leads to a vacuously true statement. The discussion emphasizes that the logical implication of "false implies false" supports this conclusion. An example provided illustrates that a relation like R = { (1, 2), (4, 3) } is transitive due to the lack of pairs to contradict the transitive property. Ultimately, the absence of complete pairs does not negate the transitive nature of the relation.
dijkarte
Messages
190
Reaction score
0
A relation R on a set S is transitive:

(x, y) and (y, x) ==> (x, z), for all pairs in R

So if I cannot find (y, z) for (x, y) in R, does this mean the relation is considered transitive since the condition still holds true because False ==> False/True evaluates to True?

Thanks.
 
Physics news on Phys.org
Fix the typo in your statement of the definition of transitive.

You aren't asking your question clearly because you aren't using any logical quantifiers on your variables (such as "there exists" or "for each". I think your question is whether a relaton on R is transitive in the case where we can't find any counterexample to it being transitive. The answer to that is yes. Your idea that this is because "false implies false" is true is basically correct.
 
OK got it, thanks!
 
dijkarte said:
A relation R on a set S is transitive:

(x, y) and (y, x) ==> (x, z), for all pairs in R

So if I cannot find (y, z) for (x, y) in R, does this mean the relation is considered transitive since the condition still holds true because False ==> False/True evaluates to True?

Thanks.

If you regard (x,y) as the antecedent and (x,z) as the consequent of a logical implication and you regard (x,y) as true and (x,z) as false), then the implication is false. However this is an unusual way to frame the concept of transitivity. Moreover,the expression should be (x,y) -> (y,z) if transitivity holds. Given that as a premise, you can say (x,y)-> (x,z)
 
Last edited:
If you regard (x,y) as the antecedent and (x,z) as the consequent of a logical implication and you regard (x,y) as true and (x,z) as false), then the implication is false. However this is an unusual way to frame the concept of transitivity. Moreover,the expression should be (x,y) -> (y,z) if transitivity holds. Given that as a premise, you can say (x,y)-> (x,z)

I have no idea what this means :)

I'm not framing the concept of transitivity, I'm trying to understand why or why not we can say a relation is transitive when there's no complete set of pairs to test the condition.

In other words R = { (1, 2), (4, 3) } is transitive, where R is a relation on the set { 1, 2, 3, 4 }, because there's no (2, a) and (3, b), so that we can check for existence of (1, a) and (4, b).
 
dijkarte said:
I have no idea what this means :)

I'm not framing the concept of transitivity, I'm trying to understand why or why not we can say a relation is transitive when there's no complete set of pairs to test the condition.

In other words R = { (1, 2), (4, 3) } is transitive, where R is a relation on the set { 1, 2, 3, 4 }, because there's no (2, a) and (3, b), so that we can check for existence of (1, a) and (4, b).

Yes, R is transitive, because as you point out, IF xRy and yRz THEN xRz. The antecedent (the IF part) is vacuously true.
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K