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So A ={2,4,7,9}

R = {(2,2), (2,4), (2,7), (2,9), (4,7), (4,9), (7,9)}

not reflexive because not all the elements from A are related to one another in R

is symmetric because 2,2 can relate to itself

is anti-symmetric because similarly, 2,2 is symmetric can give 2=2

is transitive because 2,2 can be paired with itself (2,2) (2,2) ⇒ (2,2)

Is my understanding correct?

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  • indeed not reflexive because e.g. we do not have $4R4$
  • not symmetric, because $2R4$ but not $4R2$
  • indeed antisymmetric, because for every pair $(a,b)$ that satisfies $aRb\wedge bRa$ (the pair $(2,2)$ is the only one here) we also have $a=b$.
  • indeed transitive but what you gave as reason for that is not okay. It must be checked that in all cases that we have $aRb\wedge bRc$ we also have $aRc$.
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is not reflexive because not all the elements from A are related to one another in R

Correct. Reflexivity requires all elements in A to be self related. $\forall x\in A:(x,x)\in R$.

is symmetric because 2,2 can relate to itself

Incorrect.   While $(2,2)$ is its own symmetrical pairing, Symmetry requires that all pairs that are in the relation have a corresponding symmetrical pairing. $\forall x\forall y: ((x,y)\in R \to (y,x)\in R)$

This is not so. $(2,7)$ is in the relation, but $(7,2)$ is not.

is anti-symmetric because similarly, 2,2 is symmetric can give 2=2

To be precise, anti-symmetry requires that all pairs in R have a cooresponding symmetrical pairing only if they are self-relations, such as for example $(2,2)$. $\forall x\forall y~(((x,y)\in R\land (y,x)\in R)\to x=y)$

is transitive because 2,2 can be paired with itself (2,2) (2,2) ⇒ (2,2)

Again it is for all cases. You cannot say it is so because one particular case holds. $\forall x\forall y\forall z~((x,y)\in R\land (y,z)\in R\to(x,z)\in R)$

So, therefore you must exhaustively check all possible cases to ensure no counter example exists to satisfy the negation of that; which is $\exists x\exists y\exists z~((x,y)\in R\land (y,z)\in R\land (x,z)\notin R)$

PS: no counter example exists.

Is my understanding correct?

Indications are no.

To ensure a universal statement is satisfied you must ascertain that all cases hold, rather than a single example.

On the other hand, finding a single counter example is all that is needed to prove a universal statement is not satisfied.

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  • $\begingroup$ @DeusSued All elements of the set must be self-related for the relation to be reflexive. Since it does not hold for all, you were correct: the relation is not reflexive. $\endgroup$ – Graham Kemp Nov 29 '18 at 0:29
  • $\begingroup$ So antisymmertic (1,2) (2,1) (2,2) because it has (2,2) which gives a=b but if there is (a,b) but not (b,a) it is still antisymmetric $\endgroup$ – Deus Sued Nov 29 '18 at 0:47
  • $\begingroup$ (a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) So(1,2) (2,1) then there should be (1,1) for transitive $\endgroup$ – Deus Sued Nov 29 '18 at 0:51

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