# Discrete Math

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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!