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March 27, 2017

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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, 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 - ,

    Recall 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 - ,

    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 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 - ,

    All 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 - ,

    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 it is reflexive, transitive, and symmetric. Thanks again for all your help.

  • Discrete Math - ,

    Correct 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 - ,

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

  • Discrete Math :) - ,

    You're welcome!

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