Discrete Math
posted by Laurey .
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, antisymmetric, 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
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 
Laurey
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 semion the right track? Also, for R3 a little confused. Thanks so much for your help. I really appreciate it! 
Discrete Math 
MathMate
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 
Laurey
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 
MathMate
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 
Laurey
Oh yea I meant to type R1, sorry it was a typo. Thank you for your help MathMate!

Discrete Math :) 
MathMate
You're welcome!
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