# Laurey

Popular questions and responses by Laurey
1. ## Discrete Math

Consider the following relations on R, the set of real numbers 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,

2. ## Discrete Math

Which of these relations on {0, 1, 2, 3} are equivalence relations? Justify the relation(s) that are not equivalent. R1: {(0,0), (1,1), (2,2), (3,3)} R2: {(0,0), (1,1), (1,3), (2,2), (2,3), (3,1), (3,2), (3,3)} R3: {(0,0), (0,1), (0,2), (1,0), (1,1),

3. ## Discrete Math

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

4. ## Discrete Math

Justifying your conclusions (you could also use examples in order to illustrate your results). What can you say about the sets A and B if we know that: 1. A ∪ B = A 2. A ∩ B = A Thanks for any helpful replies :)

1. ## Discrete Math

Thank you for the reassurance.

posted on February 8, 2011
2. ## Discrete Math

I think I may have found the problem in my thinking: R2 is not equivalent right? Because it is not transitive. Justification: It is reflexive because the relation does contain (0,0), (1,1), (2,2), and (3,3). It is symmetric because the relation contains

posted on February 8, 2011
3. ## Discrete Math

OooOOo. . .thank you so much for all your help.

posted on February 8, 2011
4. ## Discrete Math

So, it is not antisymmetric because 2 ≠ 3, but what would have made it true?

posted on February 8, 2011
5. ## Discrete Math

Oh yea I meant to type R1, sorry it was a typo. Thank you for your help MathMate!

posted on February 8, 2011
6. ## Discrete Math

R3: 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

posted on February 8, 2011
7. ## Discrete Math

Thank 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

posted on February 7, 2011