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)
R2: x, y ∈R iff x≥y
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.
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!
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.
Not Reflexive: x ⊀ x
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.
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)
Oh yea I meant to type R1, sorry it was a typo. Thank you for your help MathMate!