Monday

December 5, 2016
Posted by **Laurey ** on Monday, February 7, 2011 at 10:10pm.

a. R1: x, y ∈ R if and only if x = y.

b. R2: x, y ∈ R if and only if x ≥ y.

c. R3 : x, y ∈ R if and only if xy < 0.

Determine whether or not each relation is flexible, symmetric, anti-symmetric, or transitive. For each property not possessed by the relation, provide a convincing example. Summarize the results in the table below. What of these are equivalence relations?

- Discrete Math -
**MathMate**, Monday, February 7, 2011 at 10:37pmRecall the definitions of the four properties:

a ~ a. (Reflexivity)

if a ~ b then b ~ a. (Symmetry)

if a~b ∧ b~a -> a=b (antisymmetry)

if a ~ b and b ~ c then a ~ c. (Transitivity)

Example:

R2: x, y ∈R iff x≥y

reflexive: x≥x

not symmetric: x≥y -> y≥x (false)

antisymmetric: x≥y ∧ y≥x -> x=y

transitive: x≥y ∧ y≥z -> x≥z

I will leave R1 and R3 for you as an exercise

Post if you need more detailed explanations. - Discrete Math -
**Laurey**, Monday, February 7, 2011 at 11:03pmThank you MathMate for your quick reply! I think I understand it a lot better after your post, but I still feel a little fuzzy. So for R1:

Reflexive: x = x

Symmetric: x = y, then y = x

antisymmetric: x = y and y = x that implies x = y (?)

Transitive: x = y and y = z, then x = z

Am I semi-on the right track? Also, for R3 a little confused. Thanks so much for your help. I really appreciate it! - Discrete Math -
**MathMate**, Tuesday, February 8, 2011 at 12:00amAll correct. It is possible that a relation is both symmetric AND antisymmetric. No worries.

Continue with R3 and post if you are not sure. Keep up the good work. - Discrete Math -
**Laurey**, Tuesday, February 8, 2011 at 8:09amR3:

Not Reflexive: x ⊀ x

Symmetric:

Antisymmetric:

Not Transitive:

I'm not sure how to justify. . . the xy and 0 is throwing me off. . .can you separate them? If that makes any sense. . .I'm lost.

But R2 would be considered an equivalent relation because it is reflexive, transitive, and symmetric. Thanks again for all your help. - Discrete Math -
**MathMate**, Tuesday, February 8, 2011 at 9:22amCorrect about the

*requirements*for equivalent relations: reflexive, symmetric and transitive.

However, R2 is*not*symmetric, so it is not an equivalent relation. Check R1 instead.

For R3, this is how you would proceed:

R2: x, y ∈R iff xy<0

not reflexive: x*x<0 (false, x*x ≥0 ∀x)

symmetric: if xy<0 -> yx<0

not antisymmetric: if xy<0 & yx<0 -> x=y (false)

not transitive: xy<0 and yz<0 -> xz<0 (false: for example, -1*2<0 and 2*(-3)<0 -> (-1)*(-3)<0 is false) - Discrete Math -
**Laurey**, Tuesday, February 8, 2011 at 9:58amOh yea I meant to type R1, sorry it was a typo. Thank you for your help MathMate!

- Discrete Math :) -
**MathMate**, Tuesday, February 8, 2011 at 10:09amYou're welcome!