Consider the following relation on R1, the set of real numbers
R1 = {(1,1), (1,2), (2,1), (2,2), (3,3), (4,4), (3,2), (2,3)}
Determine whether or not each relation is flexible, symmetric, anti-symmetric, or transitive.
* Reflexive because the relation contains (1,1), (2,2), (3,3), and (4,4)
* Symmetric because it contains (1,2) & (2,1) and (3,2) & (2,3)
* Antisymmetric (I'm confused with this one)
* Not Transitive because it contains (2,1) & (2,3) but not (1,3)
Would this be considered correct? I'm not sure about antisymmetric. Thanks for any helpful replies
reflexive, OK.
symmetry: OK
antisymmetry:
recall: if a~b ∧ b~a -> a=b
(2,3)∧(3,2) [both true] -> 2=3 ?
not transitive: OK
(3,2) ∧ (2,1) [both true] -> (3,1) [false]
So, it is not antisymmetric because 2 ≠ 3, but what would have made it true?
Similar to a≥b!
Consider
R:{(3,2),(2,2),(3,3)}
"Antisymmetric if a~b ∧ b~a -> a=b "
(3,2)[true] ∧ (2,3)[false] -> 2=3 [true]
(because the statement is true whenever the condition is false)
(3,3)[true] ∧ (3,3)[true] -> 3=3 [true]
So R is antisymmetric (but not symmetric because (3,2) -> (2,3) [false]
OooOOo. . .thank you so much for all your help.
You're welcome!
Your analysis is mostly correct. Let's go through each property to determine whether the relation is reflexive, symmetric, antisymmetric, or transitive.
Reflexive: A relation is reflexive if every element in the set is related to itself. You correctly pointed out that the relation contains (1,1), (2,2), (3,3), and (4,4), so it is reflexive.
Symmetric: A relation is symmetric if for every (a,b) in the relation, (b,a) is also in the relation. You correctly identified (1,2) & (2,1) and (3,2) & (2,3) as pairs that satisfy this condition, so the relation is symmetric.
Antisymmetric: A relation is antisymmetric if for every (a,b) and (b,a) in the relation, where a ≠ b, it is not the case that both (a,b) and (b,a) are in the relation. In other words, if (a,b) is in the relation, then (b,a) cannot be in the relation unless a = b.
In your example, you correctly identified (1,2) and (2,1) as pairs that satisfy this condition. Since 1 ≠ 2, the presence of (1,2) implies that we cannot have (2,1) in the relation. Similarly, (2,3) and (3,2) satisfy this condition. Since 2 ≠ 3, the presence of (2,3) implies that we cannot have (3,2) in the relation.
So, the relation is antisymmetric.
Transitive: A relation is transitive if for every (a,b) and (b,c) in the relation, then (a,c) is also in the relation. In your example, you correctly pointed out that (2,1) and (2,3) are in the relation, but (1,3) is not. Since (2,1) and (2,3) are in the relation and (1,3) is not, the relation is not transitive.
To summarize:
- The relation is reflexive because it contains (1,1), (2,2), (3,3), and (4,4).
- The relation is symmetric because it contains (1,2) & (2,1) and (3,2) & (2,3).
- The relation is antisymmetric because (1,2) and (2,1) cannot both be in the relation, and (2,3) and (3,2) cannot both be in the relation.
- The relation is not transitive because it contains (2,1) & (2,3) but not (1,3).
Your analysis is correct!