Posted by **Laurey ** on Tuesday, February 8, 2011 at 9:04am.

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

- Discrete Math -
**MathMate**, Tuesday, February 8, 2011 at 10:25am
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]

- Discrete Math -
**Laurey **, Tuesday, February 8, 2011 at 10:28am
So, it is not antisymmetric because 2 ≠ 3, but what would have made it true?

- Discrete Math -
**MathMate**, Tuesday, February 8, 2011 at 10:46am
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]

- Discrete Math -
**Laurey **, Tuesday, February 8, 2011 at 11:01am
OooOOo. . .thank you so much for all your help.

- Discrete Math -
**Anonymous**, Tuesday, February 8, 2011 at 12:22pm
You're welcome!

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