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]
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
5 answers
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]
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!
3 answers