# Properties on relation (reflexive, symmetric, anti-symmetric and transitive)

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?

• 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 $$aRc$$.

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.

• @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. – Graham Kemp Nov 29 '18 at 0:29
• 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 – Deus Sued Nov 29 '18 at 0:47
• (a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) So(1,2) (2,1) then there should be (1,1) for transitive – Deus Sued Nov 29 '18 at 0:51