Proving R is a pertial order Set.

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Discussion Overview

The discussion revolves around proving that a specific relation R defined on the set X = {0, 1, 2, ..., 5} is a partial order relation. Participants explore the properties of reflexivity, antisymmetry, and transitivity required for R to qualify as a partial order.

Discussion Character

  • Homework-related, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant expresses confusion about the problem and seeks help in proving that R is a partial order.
  • Another participant notes that since X has only 6 elements, it is feasible to explore the pairs and possibly draw a partial ordering.
  • A participant emphasizes the need to prove that R is reflexive, antisymmetric, and transitive, providing a specific example for reflexivity.
  • There is a suggestion that antisymmetry can be checked by examining all pairs (x, y) where x ≠ y.
  • A later reply acknowledges a misunderstanding of the concept of partial order and expresses gratitude for the clarification.

Areas of Agreement / Disagreement

Participants generally agree on the properties that need to be proven for R to be a partial order, but the discussion remains unresolved regarding the specific proofs of antisymmetry and transitivity.

Contextual Notes

Participants have not fully explored the proofs for antisymmetry and transitivity, and there may be assumptions about the understanding of these properties that are not explicitly stated.

raf1qu3
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Ok, here is my problem. i had this problem in my Logic test last week and i felt clueless. i don't know why but I am still trying to figure it out.

here it is.

The relation R is defined on the set X= {0,1,2,..5} by the rule xRy if and only if x+2y is divisible by 3. Prove that R is a partial order relation.

i did this sort of stuff in my first yr with my eyes closed but when this came up in this years class test i was dumb founded. I have had a look at my revision notes from year 1 but they didnt help.

how would you guys prove this?
 
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Well, X only has 6 elements -- and there are only 36 pairs. It seems that (what seems to me to be) most obvious, direct thing to do -- try and draw the partial ordering -- is feasible.
 
Hurkyl said:
Well, X only has 6 elements -- and there are only 36 pairs. It seems that (what seems to me to be) most obvious, direct thing to do -- try and draw the partial ordering -- is feasible.

well for 7 marks out 30 it wouldn't be so simple. i tried drawing a digraph but how would that help?
 
What is a partial order? It's a binary relation that it's reflexive, antisymmetric and transitive, therefore you must prove that R has this properties.

For example, why is it reflexive? To be reflexive, you must prove that [tex]\forall x \in X \left(xRx\right)[/tex], but:

[tex]xRx \Leftrightarrow 3|x + 2x \Leftrightarrow 3|3x[/tex]

Which is clearly true.

For antisymmetry, you must prove that:

[tex]\forall xy \left(xRy\wedge yRx \Rightarrow x = y\right)[/tex]

But if 3|x + 2y and 3|y + 2x and this, I believe must be done by checking all pairs (x,y) such that [tex]x\neq y[/tex].

I'll leave the transitivity to you.
 
JSuarez said:
What is a partial order? It's a binary relation that it's reflexive, antisymmetric and transitive, therefore you must prove that R has this properties.

JESUS.. i can't believe i missed that... i must have had partial order mixed up with something else... thanks bud... will remember this for my end of semester exam...
this shows you must always revise the basics before you can advance, thanks again.
 

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