Laurey
Popular questions and responses by Laurey
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,
asked on February 7, 2011 
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),
asked on February 8, 2011 
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, antisymmetric, or transitive. * Reflexive because the relation
asked on February 8, 2011 
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 :)
asked on February 4, 2011

Discrete Math
Thank you for the reassurance.
posted on February 8, 2011 
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 
Discrete Math
OooOOo. . .thank you so much for all your help.
posted on February 8, 2011 
Discrete Math
So, it is not antisymmetric because 2 ≠ 3, but what would have made it true?
posted on February 8, 2011 
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 
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 
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