Is the relation reflexive, symmetric and antisymmetric?

Question

So I really want you guys to tell me if I'm wrong on any point. I will present this to my teacher and I really have to have everything well-explained, motivating myself correctly everywhere:

My task: Put A={1,2,3} and form the relation R on A by putting R = {(1,1),(2,2),(3,3),(1,2),(2,3),(3,1).

So, is the relation reflexive, symmetric, antisymmetric? Prove for every property yes or no.

So my attempt:

It is Reflexive. This is because every element in A (1,2,3) are reflected as pair in R (1,1), (2,2), (3,3).

It is not symmetric. This is because there in R exists the pairs (1,2),(2,3) and (3,1) but the pairs (2,1), (3,2) and (1,3) do not exist in R. For symmetry, for every pair of numbers in R there must be another pair of numbers thats the reverse of the other pair, like (3,4) and (4,3) for instance.

It is antisymmetric.

The definition of antisymmetry is: xRy∧yRx⇒x=y

We rewrite the rule a bit and get:

xRy ∧ yRx ⇒ x=y

¬(xRy ∧ yRx) ∨ x=y (A⇒B≡¬A∨B)

¬(xRy)∨¬(yRx)∨x=y (Duality)

¬(xRy)∨x=y∨¬(yRx) (Commutativity of ∨)

¬(xRy∧x≠y)∨¬(yRx) (Duality)

xRy ∧ x≠y⇒ ¬(yRx) (A⇒B≡¬A∨B)

so in the end we get xRy^x≠y⇒¬(yRx)

This means that whenever we have xRy but x!=y, then y never relates to x.

When we look through our pairs, this is true,

(1,2): x!=y ^ 1R2 ¬(2R1)

(2,3): x!=y ^ 2R3 ¬(3R2)

(3,1): x!y ^ 3R1 ¬(1R3)

Thus we see that for all our numbers where x!=y, we never have that both xRy and yRx at the same time. Therefore the proof of antisymmetry works and the relation is antisymmetric.

Are my answers right (I think i am)? If not, where did I get it wrong?

Greatly thankful for any help, I have to present them this week.

Answer 1

$A=\{1,2,3\}, R = \{(1,1),(2,2),(3,3),(1,2),(2,3),(3,1)\}$

The relation is reflexive because, as you say, for every $a$ in $A$, we have $(a,a)$ in $R$. $$\text{reflexivity:}~\forall a \in A~.(a,a)\in R$$

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The relation is not symmetric because there exists some pair $(a, b)$ in $R$ but the inverse, $(b,a)$ is not . Any one from the three counterexamples you found is sufficient to show this.$$\text{non-symmetry:}~\exists a\in A~\exists b\in A~.((a,b)\in R\wedge (b,a)\notin R)$$

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The relation is anti-symmetric because when any pair $(a,b)$ and its inverse $(b,a)$ are both in $R$, then the members of the pair are identical.   Remember, a conditional statement is considered true when its consequent is true or its antecedent is false ~~ it is only false when the antecedent is true but the consequent is false. $$\forall a\in A~\forall b\in A~.((a,b)\in R\wedge (b,a)\in R\to a=b)\\\forall a\in A~\forall b\in A~.(a\neq b\wedge (a,b)\in R\to(b,a)\notin R)\\\forall a\in A~\forall b\in A~.(a\neq b\to (a,b)\notin R\vee (b,a)\notin R)$$

So thence: The relation is anti-symmetric because there does not exist a counterexample where a pair and its inverse are in the relation but the members are distinct. As you found.

$$\neg\exists a\in A~\exists b\in A~.((a,b)\in R\wedge(b,a)\in R\wedge a\neq b)$$

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So 3 out of 3.

Answer 2

Reflexive and simetric I am in agreement with you. But antisimetric I argument of this form: The only pairs that verifique that xRy ^ yRx are: $(1,1),(2,2),(3,3)$ and in this three cases we have that $x=y$ so is antisimetric.